Differential equations (DE) - relationship between an unknown function y y y and its derivative
F ( y ( x ) , y ′ ( x ) , y ′ ′ ( x ) , … ) = 0 F(y(x), y'(x), y''(x), …) = 0 F ( y ( x ) , y ′ ( x ) , y ′ ′ ( x ) , … ) = 0
Order - order of the highest order derivative in the DE
Ordinary differential equation (ODE) - contains derivative with respect to only 1 variable
Partial differential equation (PDE) - contains derivative with respect to multiple variables
Linear - unknowns of DE do not appear as argument of nonlinear functions or multiply with each other or themselves
Valid solution - a solution of an ODE that does not…
have a complex number
have lim x → a = ∞ \lim\limits_{x\to a} = \infty x → a lim = ∞ (lim x → ∞ = ∞ \lim\limits_{x\to \infty} = \infty x → ∞ lim = ∞ is ok)
have division by zero
have undefined operation (e.g. ln ( − 2 ) \ln(-2) ln ( − 2 ) )
Interval of validity - the largest interval where the solution is valid.
Distinct solution - y 1 ( x ) ≠ y 2 ( x ) y_{1}(x) \not = y_{2}(x) y 1 ( x ) = y 2 ( x ) for some x x x
Existence and uniqueness theorem
1st order ODE
Consider the IVP y ′ = f ( t , y ) y' = f(t, y) y ′ = f ( t , y ) , y ( 0 ) = 0 y(0) = 0 y ( 0 ) = 0 .
If f f f and ∂ f ∂ y \frac{\partial f}{\partial y} ∂ y ∂ f are continuous for ∣ t ∣ ≤ a , ∣ y ∣ ≤ b \lvert t \rvert \le a, \lvert y \rvert \le b ∣ t ∣ ≤ a , ∣ y ∣ ≤ b ,
then there exists a ∣ t ∣ ≤ h ≤ a \lvert t \rvert \le h \le a ∣ t ∣ ≤ h ≤ a such that there exists a unique solution y ( t ) = ϕ ( t ) y(t) = \phi(t) y ( t ) = ϕ ( t ) of the IVP.
Given some initial values y ( 0 ) = a 0 , y ′ ( 0 ) = a 1 , … , y ( n ) ( 0 ) = a n y(0) = a_{0}, y'(0) = a_{1}, …, y^{(n)}(0) = a_{n} y ( 0 ) = a 0 , y ′ ( 0 ) = a 1 , … , y ( n ) ( 0 ) = a n . Solve the ODE F ( y ( x ) , y ′ ( x ) , y ′ ′ ( x ) , … ) = 0 F(y(x), y'(x), y''(x), …) = 0 F ( y ( x ) , y ′ ( x ) , y ′ ′ ( x ) , … ) = 0
Find the general solution to the ODE (or verify a given solution).
Plug in any initial values to determine the values of unknown constants in the ODE general solution.
Check if there is any additional prompt to do with the solution of the ODE.
1st order
Nonlinear, linear
Separable
Write the DE in the form of separable DE d y d x = g ( x ) h ( y ) \dfrac{dy}{dx} = g(x) h(y) d x d y = g ( x ) h ( y ) .
Check if h ( y ) = 0 h(y) = 0 h ( y ) = 0 generates trivial solutions.
Separate the x x x and y y y terms: d y h ( y ) = g ( x ) d x \dfrac{dy}{h(y)} = g(x) dx h ( y ) d y = g ( x ) d x . (Don’t consider int. const.)
Integrate both sides and add constant of integration: ∫ d y h ( y ) = ∫ g ( x ) d x \displaystyle\int\dfrac{dy}{h(y)} = \int g(x) dx ∫ h ( y ) d y = ∫ g ( x ) d x .
Solve for y y y for explicit solution.
IVP
Write the DE in the form of y ′ ( x ) + p ( x ) y ( x ) = q ( x ) y'(x) + p(x)y(x) = q(x) y ′ ( x ) + p ( x ) y ( x ) = q ( x ) .
Find the integrating factor μ ( x ) = exp ( ∫ p ( x ) d x ) \mu(x) = \exp(\int p(x)dx) μ ( x ) = exp ( ∫ p ( x ) d x ) .
The solution is in the form y ( x ) = ∫ μ ( x ) q ( x ) d x + C μ ( x ) y(x) = \dfrac{\displaystyle\int \mu(x)q(x)dx + C}{\mu(x)} y ( x ) = μ ( x ) ∫ μ ( x ) q ( x ) d x + C .
IVP
1st order
Nonlinear, linear
Exact
Write the DE in the form of N ( x , y ) y ′ + M ( x , y ) = 0 N(x, y)y' + M(x, y) = 0 N ( x , y ) y ′ + M ( x , y ) = 0
Let y = f ( x , y ) y = f(x, y) y = f ( x , y ) .
Let M ( x , y ) = ∂ f ∂ x M(x, y) = \dfrac{\partial f}{\partial x} M ( x , y ) = ∂ x ∂ f , and N ( x , y ) = ∂ f ∂ y N(x, y) = \dfrac{\partial f}{\partial y} N ( x , y ) = ∂ y ∂ f
Check if the DE is an exact DE by verifying ∂ M ∂ y = ∂ N ∂ x \dfrac{\partial M}{\partial y} = \dfrac{\partial N}{\partial x} ∂ y ∂ M = ∂ x ∂ N
Find the solution f ( x , y ) f(x, y) f ( x , y ) using one of the following methods:
Method A1
Find f ( x , y ) = ∫ ∂ f = ∫ M ∂ x = f main ( x , y ) + h ( y ) \displaystyle f(x, y) = \int \partial f = \int M \ \partial x = f_{\text{main}}(x, y) + h(y) f ( x , y ) = ∫ ∂ f = ∫ M ∂ x = f main ( x , y ) + h ( y ) .
Solve for h ′ ( y ) h'(y) h ′ ( y ) in N = ∂ f ∂ y = ∂ f main ∂ y + h ′ ( y ) N = \dfrac{\partial f}{\partial y} = \dfrac{\partial f_{\text{main}}}{\partial y} + h'(y) N = ∂ y ∂ f = ∂ y ∂ f main + h ′ ( y ) .
Solve for h ( y ) h(y) h ( y ) and plug in f ( x , y ) f(x, y) f ( x , y ) .
Method A2
Find f ( x , y ) = ∫ ∂ f = ∫ N ∂ y = f main ( x , y ) + g ( x ) \displaystyle f(x, y) = \int \partial f = \int N \ \partial y = f_{\text{main}}(x, y) + g(x) f ( x , y ) = ∫ ∂ f = ∫ N ∂ y = f main ( x , y ) + g ( x ) .
Solve for g ′ ( x ) g'(x) g ′ ( x ) in M = ∂ f ∂ x = ∂ f main ∂ x + g ′ ( x ) M = \dfrac{\partial f}{\partial x} = \dfrac{\partial f_{\text{main}}}{\partial x} + g'(x) M = ∂ x ∂ f = ∂ x ∂ f main + g ′ ( x ) .
Solve for g ( x ) g(x) g ( x ) and plug in f ( x , y ) f(x, y) f ( x , y ) .
Method B
Do not consider int. const. g ( x ) g(x) g ( x ) and h ( y ) h(y) h ( y ) in the following steps.
Find f 1 ( x , y ) ≡ ∫ ∂ f = ∫ M ∂ x ≡ f mixed-terms ( x , y ) + f y-terms ( y ) \displaystyle f_{1}(x, y) \equiv \int \partial f = \int M \ \partial x \equiv f_{\text{mixed-terms}}(x, y) + f_{\text{y-terms}}(y) f 1 ( x , y ) ≡ ∫ ∂ f = ∫ M ∂ x ≡ f mixed-terms ( x , y ) + f y-terms ( y )
Find f 2 ( x , y ) ≡ ∫ ∂ f = ∫ N ∂ y ≡ f mixed-terms ( x , y ) + f x-terms ( x ) \displaystyle f_{2}(x, y) \equiv \int \partial f = \int N \ \partial y \equiv f_{\text{mixed-terms}}(x, y) + f_{\text{x-terms}}(x) f 2 ( x , y ) ≡ ∫ ∂ f = ∫ N ∂ y ≡ f mixed-terms ( x , y ) + f x-terms ( x )
Match up the terms to get the solution f ( x , y ) = f mixed-terms ( x , y ) + f x-terms ( x ) + f y-terms ( y ) = C f(x, y) = f_{\text{mixed-terms}}(x, y) + f_{\text{x-terms}}(x) + f_{\text{y-terms}}(y) = C f ( x , y ) = f mixed-terms ( x , y ) + f x-terms ( x ) + f y-terms ( y ) = C
IVP
1st order
Nonlinear, linear
Substitute function y y y with u u u to find an easy-to-solve ODE by…
Given DE y ′ = f ( x , y ) y' = f(x, y) y ′ = f ( x , y ) ,
write the DE in terms of y ′ y' y ′
find the substitution u = u ( x , y ( x ) ) u = u(x, y(x)) u = u ( x , y ( x ) )
find the inverse substitution y = y ( x , u ( x ) ) y = y(x, u(x)) y = y ( x , u ( x ) )
By chain rule, find u ′ = ∂ u ∂ x + ∂ u ∂ y y ′ u' = \dfrac{\partial u}{\partial x} + \dfrac{\partial u}{\partial y} y' u ′ = ∂ x ∂ u + ∂ y ∂ u y ′ .
Substitution (replace y ′ y' y ′ with original ODE): u ′ = ∂ u ∂ x + ∂ u ∂ y y ′ ← y ′ y ′ = f ( x , y ) u' = \dfrac{\partial u}{\partial x} + \dfrac{\partial u}{\partial y} y' \xleftarrow{y'} y' = f(x, y) u ′ = ∂ x ∂ u + ∂ y ∂ u y ′ y ′ y ′ = f ( x , y ) u ′ = ∂ u ∂ x + ∂ u ∂ y f ( x , y ) ≡ F ( x , y , u ) u' = \dfrac{\partial u}{\partial x} + \dfrac{\partial u}{\partial y} f(x, y) \equiv F(x, y, u) u ′ = ∂ x ∂ u + ∂ y ∂ u f ( x , y ) ≡ F ( x , y , u )
Substitution (replace y y y with inverse substitution): u ′ = F ( x , y , u ) ← y y ( x , u ( x ) ) u' = F(x, y, u) \xleftarrow{y} y(x, u(x)) u ′ = F ( x , y , u ) y y ( x , u ( x ) ) u ′ = F ( x , y ( x , u ) , u ) ≡ G ( x , u ) u' = F(x, y(x, u), u) \equiv G(x, u) u ′ = F ( x , y ( x , u ) , u ) ≡ G ( x , u )
Solve the ODE u ′ = G ( x , u ) u' = G(x, u) u ′ = G ( x , u ) for u ( x ) u(x) u ( x ) .
Substitution: y ( x , u ( x ) ) ← u ( x ) u ( x ) y(x, u(x)) \xleftarrow{u(x)} u(x) y ( x , u ( x ) ) u ( x ) u ( x ) y ( x ) y(x) y ( x )
IVP
Description
Equations
DE of radioactive decay
d N d t = − k N ( t ) \dfrac{dN}{dt} = -k N(t) d t d N = − k N ( t )
Number of nuclei over time (solution)
N ( t ) = N ( 0 ) e − k t = N 0 e − k t N(t) = N(0)e^{-kt} = N_{0}e^{-kt} N ( t ) = N ( 0 ) e − k t = N 0 e − k t
Half life
N ( τ ) = N 0 2 N(\tau) = \dfrac{N_{0}}{2} N ( τ ) = 2 N 0
Decay constant
k = ln 2 τ k = \dfrac{\ln2}{\tau} k = τ ln 2
Radioactive dating
t = ln ( N 1 ( t ) N 2 ( t ) N 2 ( 0 ) N 1 ( 0 ) ) ln 2 τ 1 τ 2 τ 1 − τ 2 t = \dfrac{\ln \left( \dfrac{N_{1}(t)}{N_{2}(t)} \dfrac{N_{2}(0)}{N_{1}(0)} \right)}{\ln 2} \dfrac{\tau_{1}\tau_{2}}{\tau_{1} - \tau_{2}} t = ln 2 ln ( N 2 ( t ) N 1 ( t ) N 1 ( 0 ) N 2 ( 0 ) ) τ 1 − τ 2 τ 1 τ 2
Description
Equations
Logistic equation (r , k > 0 r, k > 0 r , k > 0 )
P ′ ( t ) = r ( 1 − P k ) P P'(t) = r \left( 1-\dfrac{P}{k} \right)P P ′ ( t ) = r ( 1 − k P ) P
Bernoulli equation
y ′ + p ( t ) y = q ( t ) y n y' + p(t)y = q(t)y^{n} y ′ + p ( t ) y = q ( t ) y n
General solutions of Bernoulli equation (n = 1 , 2 , 3 n = 1,2,3 n = 1 , 2 , 3 )
n = 0 : y ( t ) = ∫ e ∫ p ( t ) d t q ( t ) d t + C e ∫ p ( t ) d t n = 1 : y ( t ) = C e ∫ q ( t ) − p ( t ) d t n = 2 : y ( t ) = − e − ∫ p ( t ) d t ∫ e − ∫ p ( t ) d t q ( t ) d t + C n = 0: y(t) = \dfrac{\displaystyle\int e^{\int p(t) dt} q(t) dt + C}{e^{\int p(t) dt}} \newline n = 1: y(t) = Ce^{\int q(t)-p(t) \ dt} \newline n = 2: y(t) = \dfrac{-e^{-\int p(t) dt}}{\displaystyle\int e^{-\int p(t) dt} q(t) dt + C} n = 0 : y ( t ) = e ∫ p ( t ) d t ∫ e ∫ p ( t ) d t q ( t ) d t + C n = 1 : y ( t ) = C e ∫ q ( t ) − p ( t ) d t n = 2 : y ( t ) = ∫ e − ∫ p ( t ) d t q ( t ) d t + C − e − ∫ p ( t ) d t
General solutions of Bernoulli equation
∀ n : y ( t ) = ( 1 − n f ( t ) ∫ q ( t ) f ( t ) d t ) 1 / ( 1 − n ) \forall n: y(t) = \left( \dfrac{1-n}{f(t)} \displaystyle\int q(t)f(t) dt \right)^{1/(1-n)} ∀ n : y ( t ) = ( f ( t ) 1 − n ∫ q ( t ) f ( t ) d t ) 1 / ( 1 − n ) where f ( t ) = e ∫ ( 1 − n ) p ( t ) d t f(t) = e^{\int (1-n) p(t) dt} f ( t ) = e ∫ ( 1 − n ) p ( t ) d t
Riccati equation
y ′ = q 0 ( t ) + q 1 ( t ) y + q 2 ( t ) y 2 y' = q_{0}(t) + q_{1}(t)y + q_{2}(t)y^{2} y ′ = q 0 ( t ) + q 1 ( t ) y + q 2 ( t ) y 2
Solving Riccati equation known particular solution y 1 y_{1} y 1
y = y 1 + 1 v ( t ) y = y_{1} + \dfrac{1}{v(t)} y = y 1 + v ( t ) 1 , where v v v satisfies v ′ ( t ) = − ( q 1 + 2 q 2 y 1 ) v − q 2 v'(t) = -(q_{1} + 2q_{2}y_{1})v - q_{2} v ′ ( t ) = − ( q 1 + 2 q 2 y 1 ) v − q 2
General solutions of Riccati equation known particular solution y 1 y_{1} y 1
y = y 1 + μ ( t ) − ∫ μ ( t ) q 2 d t + C y = y_{1} + \dfrac{\mu(t)}{-\displaystyle\int \mu(t)q_{2} dt + C} y = y 1 + − ∫ μ ( t ) q 2 d t + C μ ( t ) where μ ( t ) = e ∫ q 1 + 2 q 2 y 1 d t \mu(t) = e^{\int q_{1}+2q_{2}y_{1} \ dt} μ ( t ) = e ∫ q 1 + 2 q 2 y 1 d t
Stability and phase plane analysis
Plot P ′ ( t ) P'(t) P ′ ( t ) vs P ( t ) P(t) P ( t ) .
Find fixed point P ∗ P^{*} P ∗ such that P ′ = 0 P' = 0 P ′ = 0
x-intercept of P ′ ( t ) P'(t) P ′ ( t ) vs P ( t ) P(t) P ( t ) plot
a solution with initial value P ( 0 ) = P ∗ P(0) = P^{*} P ( 0 ) = P ∗ is constant over time
Draw flow arrows near fixed points
P ′ > 0 P' > 0 P ′ > 0 flows to the right
P ′ < 0 P' < 0 P ′ < 0 flows to the left
Identify stability of fixed points by flow arrows
stable - solution approaches the fixed point
unstable - solution diverges from the fixed point
semi-stable - solution approaches the fixed point from one side and diverges from another
Sketch solution P ( t ) P(t) P ( t ) vs. t t t so that
P P P increases in region flow to the right
P P P decreases in region flow to the left
solution approaches to stable fixed point
solution diverges from unstable fixed point
Second order linear differential equation - r ( x ) y ′ ′ + p ( x ) y ′ + q ( x ) y = g ( x ) r(x)y'' + p(x)y' + q(x)y = g(x) r ( x ) y ′ ′ + p ( x ) y ′ + q ( x ) y = g ( x )
Homogeneous - g ( x ) = 0 g(x) = 0 g ( x ) = 0
Non-homogeneous - g ( x ) ≠ 0 g(x) \not= 0 g ( x ) = 0
Linearly independent - c 1 f ( x ) + c 2 g ( x ) = 0 c_{1}f(x) + c_{2}g(x) = 0 c 1 f ( x ) + c 2 g ( x ) = 0 can only be satisfied by choosing c 1 = c 2 = 0 c_{1} = c_{2} = 0 c 1 = c 2 = 0 for functions f , g f, g f , g
Wronskian - W ( f , g ) ( x ) = f g ′ − f ′ g W(f, g)(x) = fg' - f' g W ( f , g ) ( x ) = f g ′ − f ′ g
2nd order homogeneous ODE
Consider 2nd order homogeneous ODE r ( x ) y ′ ′ + p ( x ) y ′ + q ( x ) y = 0 r(x)y'' + p(x)y' + q(x)y = 0 r ( x ) y ′ ′ + p ( x ) y ′ + q ( x ) y = 0 .
If Wronskian W ( y 1 , y 2 ) ≠ 0 W(y_{1}, y_{2}) \not = 0 W ( y 1 , y 2 ) = 0 (y 1 y_{1} y 1 and y 2 y_{2} y 2 are linearly independent solution of the ODE),
then the general solution of the ODE is y ( x ) = c 1 y 1 ( x ) + c 2 y 2 ( x ) y(x) = c_{1}y_{1}(x) + c_{2}y_{2}(x) y ( x ) = c 1 y 1 ( x ) + c 2 y 2 ( x ) .
Wronskian and linear (in)dependence
Two functions f f f and g g g are linearly dependent
if their Wronskian W ( f , g ) ( x ) = f g ′ − f ′ g = 0 W(f, g)(x) = fg' - f' g = 0 W ( f , g ) ( x ) = f g ′ − f ′ g = 0 .
Corollary
two linearly independent functions f f f and g g g has W ( f , g ) ( x ) ≠ 0 W(f, g)(x) \not= 0 W ( f , g ) ( x ) = 0 .
If y 1 y_{1} y 1 and y 2 y_{2} y 2 be any two solutions of y ′ ′ + p ( x ) y ′ + q ( x ) y = 0 y'' + p(x)y' + q(x)y = 0 y ′ ′ + p ( x ) y ′ + q ( x ) y = 0 ,
then W ( y 1 , y 2 ) = x e − ∫ p ( d ) d x W(y_{1}, y_{2}) = xe^{-\int p(d) dx} W ( y 1 , y 2 ) = x e − ∫ p ( d ) d x .
Corollary: reduction of order formula
Known y 1 y_{1} y 1 , then y 2 = y 1 ∫ W y 1 2 d x y_{2} = y_{1}\displaystyle\int\dfrac{W}{y_{1}^{2}}dx y 2 = y 1 ∫ y 1 2 W d x
e i β x = cos ( β x ) + i sin ( β x ) e^{i\beta x} = \cos(\beta x) + i\sin(\beta x) e i β x = cos ( β x ) + i sin ( β x )
e − i β x = cos ( β x ) − i sin ( β x ) e^{-i\beta x} = \cos(\beta x) - i\sin(\beta x) e − i β x = cos ( β x ) − i sin ( β x )
2nd order
Linear
Constant coefficient
Homogeneous
Write the ODE in the form of a y ′ ′ + b y ′ + c y = 0 ay'' + by' + cy = 0 a y ′ ′ + b y ′ + c y = 0 .
Write the characteristic equation a λ 2 + b λ + c = 0 a\lambda^{2} + b\lambda + c = 0 a λ 2 + b λ + c = 0 .
Solve the characteristic equation λ 1 , λ 2 = − b ± b 2 − 4 a c 2 a \lambda_{1}, \lambda_{2} = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a} λ 1 , λ 2 = 2 a − b ± b 2 − 4 a c .
If λ 1 ≠ λ 2 \lambda_{1} \not = \lambda_{2} λ 1 = λ 2 and λ 1 , λ 2 ∈ R \lambda_{1}, \lambda_{2} \in \Reals λ 1 , λ 2 ∈ R ,
then the general solution is y ( x ) = c 1 e λ 1 x + c 2 e λ 2 x y(x) = c_{1}e^{\lambda_{1}x} + c_{2}e^{\lambda_{2}x} y ( x ) = c 1 e λ 1 x + c 2 e λ 2 x .
If λ 1 = λ 2 \lambda_{1} = \lambda_{2} λ 1 = λ 2 , and λ 1 , λ 2 ∈ R \lambda_{1}, \lambda_{2} \in \Reals λ 1 , λ 2 ∈ R ,
If λ 1 ≠ λ 2 \lambda_{1} \not = \lambda_{2} λ 1 = λ 2 , and λ 1 , λ 2 ∈ C \lambda_{1}, \lambda_{2} \in \Complex λ 1 , λ 2 ∈ C , where λ 1 , λ 2 = α ± i β \lambda_{1}, \lambda_{2} = \alpha \pm i\beta λ 1 , λ 2 = α ± i β ,
then the general solution is y ( x ) = c 1 e α x cos ( β x ) + c 2 e α x sin ( β x ) y(x) = c_{1}e^{\alpha x} \cos(\beta x) + c_{2}e^{\alpha x} \sin(\beta x) y ( x ) = c 1 e α x cos ( β x ) + c 2 e α x sin ( β x ) ,
where α = − b 2 a \alpha = -\dfrac{b}{2a} α = − 2 a b , β = 4 a c − b 2 2 a \beta = \dfrac{\sqrt{4ac-b^{2}}}{2a} β = 2 a 4 a c − b 2 .
by Euler’s formula
IVP
2nd order
Linear
Non-constant coefficient, constant coefficient
Homogeneous
Given a solution y 1 y_{1} y 1 .
Write the ODE in the form of y ′ ′ + p ( x ) y ′ + q ( x ) y = 0 y'' + p(x) y' + q(x)y = 0 y ′ ′ + p ( x ) y ′ + q ( x ) y = 0 .
Calculate the Wronskian by Abel’s Theorem W = c e − ∫ p ( x ) d x W = ce^{-\int p(x)dx} W = c e − ∫ p ( x ) d x .
Find y 2 y_{2} y 2 by reduction of order formula y 2 = y 1 ∫ W y 1 2 d x y_{2} = y_{1} \displaystyle\int \dfrac{W}{y_{1}^{2}} \ dx y 2 = y 1 ∫ y 1 2 W d x .
Pick a convenient coefficient c c c (but c ≠ 0 c \not= 0 c = 0 ).
Write the general solution y ( x ) = c 1 y 1 ( x ) + c 2 y 2 ( x ) y(x) = c_{1}y_{1}(x) + c_{2}y_{2}(x) y ( x ) = c 1 y 1 ( x ) + c 2 y 2 ( x ) .
IVP
2nd order
Linear
Non-constant coefficient (of special type)
Homogeneous
Euler equation
Write the ODE in the form of x 2 y ′ ′ + α x y ′ + β y = 0 x^{2}y'' + \alpha xy' + \beta y = 0 x 2 y ′ ′ + α x y ′ + β y = 0 .
Write the indicial equation s 2 + ( α − 1 ) s + β = 0 s^{2} + (\alpha - 1)s + \beta = 0 s 2 + ( α − 1 ) s + β = 0 .
Solve the indicial equation s 1 , s 2 = 1 − α ± ( α − 1 ) 2 − 4 β 2 s_{1}, s_{2} = \dfrac{1 - \alpha \pm \sqrt{(\alpha - 1)^{2} - 4\beta}}{2} s 1 , s 2 = 2 1 − α ± ( α − 1 ) 2 − 4 β .
If s 1 ≠ s 2 s_{1} \not = s_{2} s 1 = s 2 and s 1 , s 2 ∈ R s_{1}, s_{2} \in \Reals s 1 , s 2 ∈ R ,
then the general solution is y ( x ) = c 1 x s 1 + c 2 x s 2 y(x) = c_{1}x^{s_{1}} + c_{2}x^{s_{2}} y ( x ) = c 1 x s 1 + c 2 x s 2 .
If s 1 = s 2 s_{1} = s_{2} s 1 = s 2 , and s 1 , s 2 ∈ R s_{1}, s_{2} \in \Reals s 1 , s 2 ∈ R ,
If s 1 ≠ s 2 s_{1} \not = s_{2} s 1 = s 2 , and s 1 , s 2 ∈ C s_{1}, s_{2} \in \Complex s 1 , s 2 ∈ C , where s 1 , s 2 = η ± i μ s_{1}, s_{2} = \eta \pm i\mu s 1 , s 2 = η ± i μ ,
then the general solution is y ( x ) = c 1 x η cos ( μ ln ( x ) ) + c 2 x η sin ( μ ln ( x ) ) y(x) = c_{1}x^{\eta} \cos(\mu \ln(x)) + c_{2}x^{\eta} \sin(\mu \ln(x)) y ( x ) = c 1 x η cos ( μ ln ( x ) ) + c 2 x η sin ( μ ln ( x ) ) ,
where η = − 1 − α 2 \eta = -\dfrac{1-\alpha}{2} η = − 2 1 − α , μ = 4 β − ( α − 1 ) 2 2 \mu = \dfrac{\sqrt{4\beta-(\alpha - 1)^{2}}}{2} μ = 2 4 β − ( α − 1 ) 2 .
by Euler’s formula
IVP
2nd order
Linear
Constant coefficient
Nonhomogeneous
Write the ODE in the form of L [ y ] ≡ a y ′ ′ + b y ′ + c y = g ( x ) L[y] \equiv ay'' + by' + cy = g(x) L [ y ] ≡ a y ′ ′ + b y ′ + c y = g ( x ) .
Calculate the solution y H y_{H} y H to the homogeneous problem L [ y H ] = 0 L[y_{H}] = 0 L [ y H ] = 0 .
Guess a particular solution y P y_{P} y P to the nonhomogeneous problem L [ y H ] = g ( x ) L[y_{H}] = g(x) L [ y H ] = g ( x ) .
Substitute y P y_{P} y P into the ODE and solve for any constants.
Write the general solution y ( x ) = y H ( x ) + y P ( x ) y(x) = y_{H}(x) + y_{P}(x) y ( x ) = y H ( x ) + y P ( x ) .
IVP
Consider each term of nonhomogeneity g ( x ) g(x) g ( x ) separately.
Beware of the constants
Changes
α → A \alpha \to A α → A
β → B \beta \to B β → B
P n ( x ) → S n ( x ) P_{n}(x) \to S_{n}(x) P n ( x ) → S n ( x ) contains α i → A i , ∀ n \alpha_{i} \to A_{i}, \forall n α i → A i , ∀ n
Q m ( x ) → T m ( x ) Q_{m}(x) \to T_{m}(x) Q m ( x ) → T m ( x ) contains α i → A i , ∀ n \alpha_{i} \to A_{i}, \forall n α i → A i , ∀ n
No changes
If a term of guessed y P y_{P} y P conflicts with y H y_{H} y H , then multiply the term of guessed y P y_{P} y P (not the entire guess of y P y_{P} y P ) by x x x .
Term of Nonhomogeneity g ( x ) g(x) g ( x )
Term of guessed particular solution y P y_{P} y P
α \alpha α
A A A
α e k x \alpha e^{kx} α e k x
A e k x A e^{kx} A e k x
P n ( x ) = ∑ i = 0 n α i x i P_{n}(x) = \sum\limits_{i=0}^{n} \alpha_{i}x^{i} P n ( x ) = i = 0 ∑ n α i x i
S n ( x ) = ∑ i = 0 n A i x i S_{n}(x) = \sum\limits_{i=0}^{n} A_{i}x^{i} S n ( x ) = i = 0 ∑ n A i x i
e k x P n ( x ) e^{k x}P_{n}(x) e k x P n ( x )
e k x S n ( x ) e^{k x}S_{n}(x) e k x S n ( x )
α cos ( ω x ) + β sin ( ω x ) \alpha\cos(\omega x) + \beta\sin(\omega x) α cos ( ω x ) + β sin ( ω x )
A cos ( ω x ) + B sin ( ω x ) A\cos(\omega x) + B\sin(\omega x) A cos ( ω x ) + B sin ( ω x )
P n ( x ) cos ( ω x ) + Q m ( x ) sin ( ω x ) P_{n}(x)\cos(\omega x) + Q_{m}(x)\sin(\omega x) P n ( x ) cos ( ω x ) + Q m ( x ) sin ( ω x )
S n ( x ) cos ( ω x ) + T m ( x ) sin ( ω x ) S_{n}(x)\cos(\omega x) + T_{m}(x)\sin(\omega x) S n ( x ) cos ( ω x ) + T m ( x ) sin ( ω x )
e k x [ P n ( x ) cos ( ω x ) + Q m ( x ) sin ( ω x ) ] e^{k x}[P_{n}(x)\cos(\omega x) + Q_{m}(x)\sin(\omega x)] e k x [ P n ( x ) cos ( ω x ) + Q m ( x ) sin ( ω x ) ]
e k x [ S n ( x ) cos ( ω x ) + T m ( x ) sin ( ω x ) ] e^{k x}[S_{n}(x)\cos(\omega x) + T_{m}(x)\sin(\omega x)] e k x [ S n ( x ) cos ( ω x ) + T m ( x ) sin ( ω x ) ]
2nd order
Linear
Non-constant coefficient, constant coefficient
Nonhomogeneous
Write the ODE in the form of L [ y ] ≡ y ′ ′ + p ( x ) y ′ + r ( x ) y = g ( x ) L[y] \equiv y'' + p(x)y' + r(x)y = g(x) L [ y ] ≡ y ′ ′ + p ( x ) y ′ + r ( x ) y = g ( x ) .
Calculate the solution to the homogeneous problem L [ y H ] = 0 L[y_{H}] = 0 L [ y H ] = 0
y H = c 1 y 1 + c 2 y 2 y_{H} = c_{1}y_{1} + c_{2}y_{2} y H = c 1 y 1 + c 2 y 2
Calculate the Wronskian W ( y 1 , y 2 ) W(y_{1}, y_{2}) W ( y 1 , y 2 )
Calculate the particular solution y P y_{P} y P to the nonhomogeneous problem L [ y H ] = g ( x ) L[y_{H}] = g(x) L [ y H ] = g ( x )
y P = − y 1 ∫ g ( x ) y 2 W ( y 1 , y 2 ) d x + y 2 ∫ g ( x ) y 1 W ( y 1 , y 2 ) d x y_{P} = -y_{1} \displaystyle\int \dfrac{g(x)y_{2}}{W(y_{1}, y_{2})} dx + y_{2} \int \dfrac{g(x)y_{1}}{W(y_{1}, y_{2})} dx y P = − y 1 ∫ W ( y 1 , y 2 ) g ( x ) y 2 d x + y 2 ∫ W ( y 1 , y 2 ) g ( x ) y 1 d x
Write the general solution y ( x ) = y H ( x ) + y P ( x ) y(x) = y_{H}(x) + y_{P}(x) y ( x ) = y H ( x ) + y P ( x ) .
IVP
A mass on a spring moves vertically in a fluid bath on Earth.
Newton’s second law adds up all the forces
∑ F = m x ′ ′ ( t ) \sum F = mx''(t) ∑ F = m x ′ ′ ( t )
F d a m p e r = − γ x ′ ( t ) F_{\mathrm{damper}} = -\gamma x'(t) F d a m p e r = − γ x ′ ( t )
F s p r i n g = − k x ( t ) F_{\mathrm{spring}} = -kx(t) F s p r i n g = − k x ( t )
F e x t e r n a l F_{\mathrm{external}} F e x t e r n a l
ODE: m x ′ ′ ( t ) + γ x ′ ( t ) + k x ( t ) = F e x t ( t ) mx''(t) + \gamma x'(t) + kx(t) = F_{\mathrm{ext}}(t) m x ′ ′ ( t ) + γ x ′ ( t ) + k x ( t ) = F e x t ( t )
No external force on the system: F e x t ( t ) ≡ 0 F_{\mathrm{ext}}(t) \equiv 0 F e x t ( t ) ≡ 0
Homogeneous ODE: m x ′ ′ + γ x ′ + k x = 0 ( m > 0 , γ , k ≥ 0 ) \boxed{mx'' + \gamma x' + kx = 0} \ (m > 0, \gamma, k \ge 0) m x ′ ′ + γ x ′ + k x = 0 ( m > 0 , γ , k ≥ 0 )
Overdamped system
γ 2 − 4 m k > 0 \gamma^{2} - 4mk > 0 γ 2 − 4 m k > 0
λ 1 ≠ λ 2 ∈ R \lambda_{1} \not= \lambda_{2} \in \Reals λ 1 = λ 2 ∈ R
General solution: x ( t ) = c 1 e λ 1 t + c 2 e λ 2 t x(t) = c_{1}e^{\lambda_{1}t} + c_{2}e^{\lambda_{2}t} x ( t ) = c 1 e λ 1 t + c 2 e λ 2 t
Critically damped system
γ 2 − 4 m k = 0 \gamma^{2} - 4mk = 0 γ 2 − 4 m k = 0
λ 1 = λ 2 ∈ R \lambda_{1} = \lambda_{2} \in \Reals λ 1 = λ 2 ∈ R
General solution: x ( t ) = c 1 e λ 1 t + c 2 t e λ 2 t x(t) = c_{1}e^{\lambda_{1}t} + c_{2}te^{\lambda_{2}t} x ( t ) = c 1 e λ 1 t + c 2 t e λ 2 t
Underdamped system
γ 2 − 4 m k < 0 \gamma^{2} - 4mk < 0 γ 2 − 4 m k < 0
λ 1 ≠ λ 2 ∈ C \lambda_{1} \not= \lambda_{2} \in \Complex λ 1 = λ 2 ∈ C
General solution: x ( t ) = e − γ t / 2 m [ c 1 cos ( ω t ) + c 2 sin ( ω t ) ] x(t) = e^{-\gamma t/2m} [c_{1} \cos(\omega t) + c_{2}\sin(\omega t)] x ( t ) = e − γ t / 2 m [ c 1 cos ( ω t ) + c 2 sin ( ω t ) ]
Undamped spring
γ = 0 \gamma = 0 γ = 0
General solution: x ( t ) = c 1 cos ( ω t ) + c 2 sin ( ω t ) x(t) = c_{1} \cos(\omega t) + c_{2}\sin(\omega t) x ( t ) = c 1 cos ( ω t ) + c 2 sin ( ω t )
Phase-amplitude form: x ( t ) = A cos ( ω t − φ ) x(t) = A\cos(\omega t - \varphi) x ( t ) = A cos ( ω t − φ )
amplitude A = c 1 2 + c 2 2 A = \sqrt{c_{1}^{2}+c_{2}^{2}} A = c 1 2 + c 2 2
phase φ = arctan ( c 2 / c 1 ) \varphi = \arctan(c_{2}/c_{1}) φ = arctan ( c 2 / c 1 )
natural frequency ω = k m \omega = \sqrt{\dfrac{k}{m}} ω = m k
period T = 2 π ω T = \dfrac{2\pi}{\omega} T = ω 2 π
Graph: oscillating wave with constant amplitude
Underdamped spring
γ > 0 \gamma > 0 γ > 0
General solution: x ( t ) = e − γ t / 2 m [ c 1 cos ( ω t ) + c 2 sin ( ω t ) ] x(t) = e^{-\gamma t/2m} [c_{1} \cos(\omega t) + c_{2}\sin(\omega t)] x ( t ) = e − γ t / 2 m [ c 1 cos ( ω t ) + c 2 sin ( ω t ) ]
Phase-amplitude form: x ( t ) = A e − γ t / 2 m cos ( ω t − φ ) x(t) = Ae^{-\gamma t/2m}\cos(\omega t - \varphi) x ( t ) = A e − γ t / 2 m cos ( ω t − φ )
amplitude A = c 1 2 + c 2 2 A = \sqrt{c_{1}^{2}+c_{2}^{2}} A = c 1 2 + c 2 2
phase φ = arctan ( c 2 / c 1 ) \varphi = \arctan(c_{2}/c_{1}) φ = arctan ( c 2 / c 1 )
natural frequency ω = k m \omega = \sqrt{\dfrac{k}{m}} ω = m k
period T = 2 π ω T = \dfrac{2\pi}{\omega} T = ω 2 π
Graph: oscillating wave with exponentially decreasing amplitude
Has external force on the system: F e x t ( t ) ≠ 0 F_{\mathrm{ext}}(t) \not= 0 F e x t ( t ) = 0
Investigate a special case of oscillating external force F e x t ( t ) ≡ F 0 cos ( Ω t ) F_{\mathrm{ext}}(t) \equiv F_{0}\cos(\Omega t) F e x t ( t ) ≡ F 0 cos ( Ω t )
Non-homogeneous ODE: m x ′ ′ + γ x ′ + k x = F 0 cos ( Ω t ) ( m , γ , k ≥ 0 ) \boxed{mx'' + \gamma x' + kx = F_{0}\cos(\Omega t)} \ (m, \gamma, k \ge 0) m x ′ ′ + γ x ′ + k x = F 0 cos ( Ω t ) ( m , γ , k ≥ 0 )
No damping, no resonance
γ = 0 , Ω ≠ ω 0 = k m \gamma = 0, \Omega \not= \omega_{0} = \sqrt{\dfrac{k}{m}} γ = 0 , Ω = ω 0 = m k
General solution: x ( t ) = ( c 1 + F 0 m ( ω 0 2 − Ω 0 2 ) ) cos ( ω 0 t ) + c 2 sin ( ω 0 t ) x(t) = \left( c_{1} + \dfrac{F_{0}}{m(\omega_{0}^{2} - \Omega_{0}^{2})} \right) \cos(\omega_{0}t) + c_{2}\sin(\omega_{0} t) x ( t ) = ( c 1 + m ( ω 0 2 − Ω 0 2 ) F 0 ) cos ( ω 0 t ) + c 2 sin ( ω 0 t )
Graph: modulated wave + beats pattern
No damping, with resonance
γ = 0 , Ω = ω 0 = k m \gamma = 0, \Omega = \omega_{0} = \sqrt{\dfrac{k}{m}} γ = 0 , Ω = ω 0 = m k
General solution: x ( t ) = c 1 cos ( ω 0 t ) + ( c 2 + F 0 2 ω 0 m t ) sin ( ω 0 t ) x(t) = c_{1}\cos(\omega_{0}t) + \left( c_{2} + \dfrac{F_{0}}{2\omega_{0}m}t \right) \sin(\omega_{0}t) x ( t ) = c 1 cos ( ω 0 t ) + ( c 2 + 2 ω 0 m F 0 t ) sin ( ω 0 t )
Graph: oscillating wave with linearly increasing amplitude
Linear combination - x = c 1 x 1 + c 2 x 2 + … + c n x n \mathbf{x} = c_1\mathbf{x}_1 + c_2\mathbf{x}_2 + … + c_n\mathbf{x}_n x = c 1 x 1 + c 2 x 2 + … + c n x n
Linearly dependent - vectors satisfy the equation c 1 x 1 + c 2 x 2 + … + c n x n = 0 c_1\mathbf{x}_1 + c_2\mathbf{x}_2 + … + c_n\mathbf{x}_n = \mathbf{0} c 1 x 1 + c 2 x 2 + … + c n x n = 0 such that the constants are not all zero
Linearly independent - vectors that are not linearly dependent
Wronskian - W [ x 1 , x 2 , … , x n ] = det ( [ x 1 , x 2 , … , x n ] ) = det ( X ) W[\mathbf{x}_1, \mathbf{x}_2, …, \mathbf{x}_n] = \det([\mathbf{x}_1, \mathbf{x}_2, …, \mathbf{x}_n]) = \det(X) W [ x 1 , x 2 , … , x n ] = det ( [ x 1 , x 2 , … , x n ] ) = det ( X )
Checking linear independence
If W [ x 1 , x 2 , … , x n ] ≠ 0 W[\mathbf{x}_1, \mathbf{x}_2, …, \mathbf{x}_n] \not= 0 W [ x 1 , x 2 , … , x n ] = 0
then they are linearly independent
Inverse of a square matrix - A − 1 A^{-1} A − 1 thats satisfies A A − 1 = A − 1 A = I n AA^{-1} = A^{-1}A = I_n A A − 1 = A − 1 A = I n
Finding matrix inverse
A = [ a b c d ] , A − 1 = 1 a d − b c [ d − b − c a ] A = \begin{bmatrix} a & b \cr c & d \end{bmatrix}, A^{-1} = \dfrac{1}{ad-bc} \begin{bmatrix} d & -b \cr -c & a \end{bmatrix} A = [ a c b d ] , A − 1 = a d − b c 1 [ d − c − b a ]
Solving systems of equations using inverse
A x = b A\mathbf{x} = \mathbf{b} A x = b
x = A − 1 b \mathbf{x} = A^{-1}\mathbf{b} x = A − 1 b
Finding matrix determinant
A = [ a b c d ] , det ( A ) = a d − b c A = \begin{bmatrix} a & b \cr c & d \end{bmatrix}, \det(A) = ad-bc A = [ a c b d ] , det ( A ) = a d − b c
A = [ a b c d e f g h i ] , det ( A ) = a ∣ e f h i ∣ − b ∣ d f g i ∣ + c ∣ d e g h ∣ A = \begin{bmatrix} a & b & c \cr d & e & f \cr g & h & i \end{bmatrix}, \det(A) = a\begin{vmatrix} e & f \cr h & i \end{vmatrix} - b\begin{vmatrix} d & f \cr g & i \end{vmatrix} + c\begin{vmatrix} d & e \cr g & h \end{vmatrix} A = ⎣ ⎢ ⎡ a d g b e h c f i ⎦ ⎥ ⎤ , det ( A ) = a ∣ ∣ ∣ ∣ ∣ e h f i ∣ ∣ ∣ ∣ ∣ − b ∣ ∣ ∣ ∣ ∣ d g f i ∣ ∣ ∣ ∣ ∣ + c ∣ ∣ ∣ ∣ ∣ d g e h ∣ ∣ ∣ ∣ ∣
Singular matrix - matrix with a determinant of 0
Equivalent statements
det ( A ) = 0 \det(A)=0 det ( A ) = 0
A A A is singular
A − 1 A^{-1} A − 1 does not exist
A x = b A\mathbf{x} = \mathbf{b} A x = b has either no solution or infinitely many solutions
columns of A A A are linearly dependent
rows of A A A are linearly dependent
Eigenvalues and eigenvectors of matrix
Eigenvector - vector v \mathbf{v} v such that A v = λ v A\mathbf{v} = \lambda\mathbf{v} A v = λ v for square matrix A A A
0 \mathbf{0} 0 is not an eigenvector by convention
Eigenvalue - constant λ \lambda λ corresponding to the eigenvector v \mathbf{v} v
Finding eigenvalues and eigenvectors
Solve for λ \lambda λ in det ( A − λ I ) = 0 \det(A-\lambda I) = 0 det ( A − λ I ) = 0
Substitute λ \lambda λ into A v = λ v A\mathbf{v} = \lambda\mathbf{v} A v = λ v to find relationship between components of eigenvectors
Define n n n auxiliary variables y 1 y_1 y 1 , …, y n y_n y n for n n n th order ODE
Let y 1 y_1 y 1 be the original function in the ODE
Let y 2 = y 1 ′ y_2 = y_1' y 2 = y 1 ′
…
Let y n = y n − 1 ′ y_n = y_{n-1}' y n = y n − 1 ′
Rearrange the ODE to isolate the highest order derivative, and write it in terms of the auxiliary variables.
Write a system of ODEs with derivatives of auxiliary variables on the left hand side and their expression on the right hand side in terms of the auxiliary variables
y 1 ′ = y 2 y_1' = y_2 y 1 ′ = y 2 (by definition)
…
y n − 1 ′ = y n y_{n-1}' = y_{n} y n − 1 ′ = y n (by definition)
y n ′ = y_{n}' = y n ′ = highest order derivative in step 2
{ y 1 ′ = a 11 y 1 + a 12 y 2 + … + a 1 n y n + b 1 y 2 ′ = a 21 y 1 + a 22 y 2 + … + a 2 n y n + b 2 ⋮ y n ′ = a n 1 y 1 + a n 2 y 2 + … + a n n y n + b n ⇒ y ′ = A y + b \begin{cases} y_1' = a_{11}y_1 + a_{12}y_2 + … + a_{1n}y_{n} + b_1 \cr y_2' = a_{21}y_1 + a_{22}y_2 + … + a_{2n}y_{n} + b_2 \cr \vdots \cr y_n' = a_{n1}y_1 + a_{n2}y_2 + … + a_{nn}y_{n} + b_n \end{cases} \Rightarrow \boxed{\mathbf{y}' = A\mathbf{y} + \mathbf{b}} ⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎧ y 1 ′ = a 1 1 y 1 + a 1 2 y 2 + … + a 1 n y n + b 1 y 2 ′ = a 2 1 y 1 + a 2 2 y 2 + … + a 2 n y n + b 2 ⋮ y n ′ = a n 1 y 1 + a n 2 y 2 + … + a n n y n + b n ⇒ y ′ = A y + b
where y = [ y 1 y 2 ⋮ y n ] , y ′ = [ y 1 ′ y 2 ′ ⋮ y n ′ ] , b = [ b 1 b 2 ⋮ b n ] , A = [ a 11 a 12 ⋯ a 1 n a 21 a 22 ⋯ a 2 n ⋮ ⋮ ⋱ ⋮ a n 1 a n 2 ⋯ a n n ] \mathbf{y} = \begin{bmatrix} y_1 \cr y_2 \cr \vdots \cr y_n \end{bmatrix}, \mathbf{y}' = \begin{bmatrix} y_1' \cr y_2' \cr \vdots \cr y_n' \end{bmatrix}, \mathbf{b} = \begin{bmatrix} b_1 \cr b_2 \cr \vdots \cr b_n \end{bmatrix}, A = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \cr a_{21} & a_{22} & \cdots & a_{2n} \cr \vdots & \vdots & \ddots & \vdots \cr a_{n1} & a_{n2} & \cdots & a_{nn} \end{bmatrix} y = ⎣ ⎢ ⎢ ⎢ ⎢ ⎡ y 1 y 2 ⋮ y n ⎦ ⎥ ⎥ ⎥ ⎥ ⎤ , y ′ = ⎣ ⎢ ⎢ ⎢ ⎢ ⎡ y 1 ′ y 2 ′ ⋮ y n ′ ⎦ ⎥ ⎥ ⎥ ⎥ ⎤ , b = ⎣ ⎢ ⎢ ⎢ ⎢ ⎡ b 1 b 2 ⋮ b n ⎦ ⎥ ⎥ ⎥ ⎥ ⎤ , A = ⎣ ⎢ ⎢ ⎢ ⎢ ⎡ a 1 1 a 2 1 ⋮ a n 1 a 1 2 a 2 2 ⋮ a n 2 ⋯ ⋯ ⋱ ⋯ a 1 n a 2 n ⋮ a n n ⎦ ⎥ ⎥ ⎥ ⎥ ⎤
Homogeneous - b = 0 \mathbf{b} = \mathbf{0} b = 0
Nonhomogeneous - b ≠ 0 \mathbf{b} \not= \mathbf{0} b = 0
Superposition principle
If the vectors x 1 , x 2 , … , x n \mathbf{x}_1, \mathbf{x}_2, …, \mathbf{x}_n x 1 , x 2 , … , x n are linearly independent solutions of the homogeneous system x ′ = P x \mathbf{x}' = P\mathbf{x} x ′ = P x
then the general solution x \mathbf{x} x is the linear combination of them x = c 1 x 1 + c 2 x 2 + … + c n x n \mathbf{x} = c_1\mathbf{x}_1 + c_2\mathbf{x}_2 + … + c_n\mathbf{x}_n x = c 1 x 1 + c 2 x 2 + … + c n x n
ODE system
1st order
Linear
Constant coefficient
Homogeneous
Write the system of ODEs in the form of x ′ = P x \mathbf{x}' = P\mathbf{x} x ′ = P x
For P v = λ v P\mathbf{v} = \lambda\mathbf{v} P v = λ v , find the eigenvalues of λ \lambda λ by solving det ( P − λ I n ) = 0 \det(P-\lambda I_n) = 0 det ( P − λ I n ) = 0
For P v = λ v P\mathbf{v} = \lambda\mathbf{v} P v = λ v , find the eigenvectors of by substitution of λ \lambda λ
Write the general solution
≤ n \le n ≤ n distinct λ ∈ R \lambda\in\R λ ∈ R ; n n n distinct real v \mathbf{v} v
General solution: x = c 1 v 1 e λ 1 t + … + c n v n e λ n t \mathbf{x} = c_1\mathbf{v}_1 e^{\lambda_1 t} + … + c_n\mathbf{v}_n e^{\lambda_n t} x = c 1 v 1 e λ 1 t + … + c n v n e λ n t
≤ n \le n ≤ n distinct λ ∈ C \lambda\in\Complex λ ∈ C ; n n n distinct complex v \mathbf{v} v
Known: x 1 = c 1 v 1 e λ 1 t = R e ( x 1 ) + i I m ( x 1 ) \mathbf{x}_1 = c_1\mathbf{v}_1 e^{\lambda_1 t} = \mathrm{Re}(\mathbf{x}_1) + i\mathrm{Im}(\mathbf{x}_1) x 1 = c 1 v 1 e λ 1 t = R e ( x 1 ) + i I m ( x 1 )
General solution: x = c 1 R e ( x 1 ) + c 2 I m ( x 1 ) \mathbf{x} = c_1 \mathrm{Re}(\mathbf{x}_1) + c_2 \mathrm{Im}(\mathbf{x}_1) x = c 1 R e ( x 1 ) + c 2 I m ( x 1 )
≤ n \le n ≤ n distinct λ \lambda λ ; < n <n < n distinct v \mathbf{v} v
General solution: x = c 1 v e λ t + c 2 ( v t e λ t + η ⃗ e λ t ) \mathbf{x} = c_1 \mathbf{v} e^{\lambda t} + c_2(\mathbf{v}te^{\lambda t} + \vec{\eta}e^{\lambda t}) x = c 1 v e λ t + c 2 ( v t e λ t + η e λ t )
Find η ⃗ \vec{\eta} η (relationship between its components) by substituting into the ODE
Linear
Non-constant coefficient (of special type)
Homogeneous
Euler system
Write the system of ODEs in the form of x ′ = P x \mathbf{x}' = P\mathbf{x} x ′ = P x
For P v = λ v P\mathbf{v} = \lambda\mathbf{v} P v = λ v , find the eigenvalues of λ \lambda λ by solving det ( P − λ I n ) = 0 \det(P-\lambda I_n) = 0 det ( P − λ I n ) = 0
For P v = λ v P\mathbf{v} = \lambda\mathbf{v} P v = λ v , find the eigenvectors of by substitution of λ \lambda λ
Write the general solution
≤ n \le n ≤ n distinct λ ∈ R \lambda\in\R λ ∈ R ; n n n distinct real v \mathbf{v} v
General solution: x = c 1 v 1 t λ 1 + … + c n v n t λ n \mathbf{x} = c_1\mathbf{v}_1 t^{\lambda_1} + … + c_n\mathbf{v}_n t^{\lambda_n} x = c 1 v 1 t λ 1 + … + c n v n t λ n
Other conditions are not discussed
Laplace transform - L [ f ( t ) ] = F ( s ) = ∫ 0 ∞ e − s t f ( t ) d t \mathcal{L}[f(t)] = F(s) = \displaystyle\int_{0}^{\infin} e^{-st}f(t) \ dt L [ f ( t ) ] = F ( s ) = ∫ 0 ∞ e − s t f ( t ) d t
Heaviside function (unit step function)
u c ( t ) = u ( t − c ) = { 0 t < c 1 t ≥ c u_c(t) = u(t-c) = \begin{cases} 0 & t < c \cr 1 & t \ge c \end{cases} u c ( t ) = u ( t − c ) = { 0 1 t < c t ≥ c
Laplace transform is linear
L [ c 1 f ( t ) + c 2 g ( t ) ] = c 1 L [ f ( t ) ] + c 2 L [ g ( t ) ] \mathcal{L}[c_{1}f(t)+c_{2}g(t)] = c_{1}\mathcal{L}[f(t)] + c_{2}\mathcal{L}[g(t)] L [ c 1 f ( t ) + c 2 g ( t ) ] = c 1 L [ f ( t ) ] + c 2 L [ g ( t ) ]
Laplace transforms of derivatives incorporate initial conditions
L [ f ′ ( t ) ] = s F ( s ) − f ( 0 ) \mathcal{L}[f'(t)] = sF(s) - f(0) L [ f ′ ( t ) ] = s F ( s ) − f ( 0 )
L [ f ′ ′ ( t ) ] = s 2 F ( s ) − s f ( 0 ) − f ′ ( 0 ) \mathcal{L}[f''(t)] = s^2F(s)-sf(0)-f'(0) L [ f ′ ′ ( t ) ] = s 2 F ( s ) − s f ( 0 ) − f ′ ( 0 )
L [ f ( n ) ( t ) ] = s n F ( s ) − s n − 1 f ( 0 ) − s n − 2 f ( 1 ) ( 0 ) − … − f ( n − 1 ) ( 0 ) \mathcal{L}[f^{(n)}(t)] = s^nF(s)-s^{n-1}f(0) - s^{n-2}f^{(1)}(0) - … - f^{(n-1)}(0) L [ f ( n ) ( t ) ] = s n F ( s ) − s n − 1 f ( 0 ) − s n − 2 f ( 1 ) ( 0 ) − … − f ( n − 1 ) ( 0 )
Heaviside function has a simple Laplace transforms
L [ u c ( t ) ] = e − s c s \mathcal{L}[u_c(t)] = \dfrac{e^{-sc}}{s} L [ u c ( t ) ] = s e − s c
Time domain translation
L [ f ( t − c ) u c ( t ) ] = e − s c L [ f ( t ) ] \mathcal{L}[f(t-c)u_{c}(t)] = e^{-sc}\mathcal{L}[f(t)] L [ f ( t − c ) u c ( t ) ] = e − s c L [ f ( t ) ]
L − 1 [ e − s c L [ f ( t ) ] ] = f ( t − c ) u c ( t ) \mathcal{L}^{-1}[e^{-sc}\mathcal{L}[f(t)]] = f(t-c)u_c(t) L − 1 [ e − s c L [ f ( t ) ] ] = f ( t − c ) u c ( t )
Laplace domain translation
L [ e c t f ( t ) ] = F ( s − c ) \mathcal{L}[e^{ct}f(t)] = F(s-c) L [ e c t f ( t ) ] = F ( s − c )
L − 1 [ F ( s − c ) ] = e c t f ( t ) \mathcal{L}^{-1}[F(s-c)] = e^{ct}f(t) L − 1 [ F ( s − c ) ] = e c t f ( t )
Time domain: difficult ODE
Laplace transform (t → s t \to s t → s )
Laplace domain: easy algebra problem
Solve the algebra problem
Laplace domain: solution to algebra problem
Inverse Laplace transform (s → t s \to t s → t )
Time domain: solution of difficult ODE
Inverse L.T. f ( t ) f(t) f ( t )
Laplace Transform F ( s ) F(s) F ( s )
Inverse L.T. f ( t ) f(t) f ( t )
Laplace Transform F ( s ) F(s) F ( s )
1 1 1
1 s \dfrac{1}{s} s 1
e a t e^{at} e a t
1 s − a \dfrac{1}{s-a} s − a 1
t n t^{n} t n
n ! s n + 1 \dfrac{n!}{s^{n+1}} s n + 1 n !
t \sqrt{t} t
π 2 s 3 / 2 \dfrac{\sqrt{\pi}}{2s^{3/2}} 2 s 3 / 2 π
sin ( a t ) \sin(at) sin ( a t )
a s 2 + a 2 \dfrac{a}{s^{2}+a^{2}} s 2 + a 2 a
t sin ( a t ) t\sin(at) t sin ( a t )
2 a s ( s 2 + a 2 ) 2 \dfrac{2as}{(s^{2}+a^{2})^{2}} ( s 2 + a 2 ) 2 2 a s
cos ( a t ) \cos(at) cos ( a t )
s s 2 + a 2 \dfrac{s}{s^{2}+a^{2}} s 2 + a 2 s
t cos ( a t ) t\cos(at) t cos ( a t )
s 2 − a 2 ( s 2 + a 2 ) 2 \dfrac{s^{2}-a^{2}}{(s^{2}+a^{2})^{2}} ( s 2 + a 2 ) 2 s 2 − a 2
sin ( a t ) − a t cos ( a t ) \sin(at)-at\cos(at) sin ( a t ) − a t cos ( a t )
2 a 3 ( s 2 + a 2 ) 2 \dfrac{2a^{3}}{(s^{2}+a^{2})^{2}} ( s 2 + a 2 ) 2 2 a 3
cos ( a t ) − a t sin ( a t ) \cos(at)-at\sin(at) cos ( a t ) − a t sin ( a t )
s ( s 2 − a 2 ) ( s 2 + a 2 ) 2 \dfrac{s(s^{2}-a^{2})}{(s^{2}+a^{2})^{2}} ( s 2 + a 2 ) 2 s ( s 2 − a 2 )
sin ( a t ) + a t cos ( a t ) \sin(at)+at\cos(at) sin ( a t ) + a t cos ( a t )
2 a s 2 ( s 2 + a 2 ) 2 \dfrac{2as^{2}}{(s^{2}+a^{2})^{2}} ( s 2 + a 2 ) 2 2 a s 2
cos ( a t ) + a t sin ( a t ) \cos(at)+at\sin(at) cos ( a t ) + a t sin ( a t )
s ( s 2 + 3 a 2 ) ( s 2 + a 2 ) 2 \dfrac{s(s^{2}+3a^{2})}{(s^{2}+a^{2})^{2}} ( s 2 + a 2 ) 2 s ( s 2 + 3 a 2 )
sinh ( a t ) \sinh(at) sinh ( a t )
a s 2 − a 2 \dfrac{a}{s^{2}-a^{2}} s 2 − a 2 a
sin ( a t + b ) \sin(at+b) sin ( a t + b )
s sin ( b ) + a cos ( b ) s 2 + a 2 \dfrac{s\sin(b)+a\cos(b)}{s^2+a^2} s 2 + a 2 s sin ( b ) + a cos ( b )
cosh ( a t ) \cosh(at) cosh ( a t )
s s 2 − a 2 \dfrac{s}{s^{2}-a^{2}} s 2 − a 2 s
cos ( a t + b ) \cos(at+b) cos ( a t + b )
s cos ( b ) − a sin ( b ) s 2 + a 2 \dfrac{s\cos(b)-a\sin(b)}{s^2+a^2} s 2 + a 2 s cos ( b ) − a sin ( b )
e a t sin ( b t ) e^{at}\sin(bt) e a t sin ( b t )
b ( s − a ) 2 + b 2 \dfrac{b}{(s-a)^2+b^2} ( s − a ) 2 + b 2 b
e a t sinh ( b t ) e^{at}\sinh(bt) e a t sinh ( b t )
b ( s − a ) 2 − b 2 \dfrac{b}{(s-a)^2-b^2} ( s − a ) 2 − b 2 b
e a t cos ( b t ) e^{at}\cos(bt) e a t cos ( b t )
s − a ( s − a ) 2 + b 2 \dfrac{s-a}{(s-a)^2+b^2} ( s − a ) 2 + b 2 s − a
e a t cosh ( b t ) e^{at}\cosh(bt) e a t cosh ( b t )
s − a ( s − a ) 2 − b 2 \dfrac{s-a}{(s-a)^2-b^2} ( s − a ) 2 − b 2 s − a
u c ( t ) u_{c}(t) u c ( t )
e − s c s \dfrac{e^{-sc}}{s} s e − s c