7.5 Undetermined coefficients
Undetermined coefficients
If a nonhomogeneous linear second-order ODE has constant coefficients
y ′ ′ + a y ′ + b y = r ( x ) y'' + a y' + b y = r(x) y ′′ + a y ′ + b y = r ( x ) we can “guess” a solution as follows:
r r r y p y_{\rm p} y p k e γ x k e^{\gamma x} k e γ x k e γ x k e^{\gamma x} k e γ x k x n ( n = 0 , 1 , … ) k x^n (n=0,1,\ldots) k x n ( n = 0 , 1 , … ) k n x n + k n − 1 x n − 1 + ⋯ + k 1 x + k 0 k_n x^n + k_{n-1} x^{n-1} + \cdots + k_1 x + k_0 k n x n + k n − 1 x n − 1 + ⋯ + k 1 x + k 0 k e α x cos ω x k e^{\alpha x} \cos{\omega x} k e αx cos ω x e α x ( k 1 cos ω x + k 2 sin ω x ) e^{\alpha x}(k_1 \cos{\omega x} + k_2 \sin{\omega x}) e αx ( k 1 cos ω x + k 2 sin ω x ) k e α x sin ω x k e^{\alpha x} \sin{\omega x} k e αx sin ω x e α x ( k 1 cos ω x + k 2 sin ω x ) e^{\alpha x}(k_1 \cos{\omega x} + k_2 \sin{\omega x}) e αx ( k 1 cos ω x + k 2 sin ω x )
If r has multiple terms, add guesses for each together.
If the guess for y p y_{\rm p} y p y h y_{\rm h} y h x unless this solution is a repeated root - then multiply by x 2 x^2 x 2
For example, to solve:
y ′ ′ + y = 0.001 x 2 , y ( 0 ) = 0 , y ′ ( 0 ) = 1.5 y'' + y = 0.001 x^2, \quad y(0) = 0, \quad y'(0) = 1.5 y ′′ + y = 0.001 x 2 , y ( 0 ) = 0 , y ′ ( 0 ) = 1.5 Use the following steps:
Find y h y_{\rm h} y h
y h ′ ′ + y h = 0 λ 2 + 1 = 0 \begin{align}
y_{\rm h}'' + y_{\rm h} &= 0 \\
\lambda^2 + 1 &= 0
\end{align} y h ′′ + y h λ 2 + 1 = 0 = 0 so λ = ± i \lambda = \pm i λ = ± i
y h = c 1 cos x + c 2 sin x y_{\rm h} = c_1 \cos x +c_2 \sin x y h = c 1 cos x + c 2 sin x Guess y p y_{\rm p} y p
y p = k 2 x 2 + k 1 x + k 0 y p ′ = 2 k 2 x + k 1 y p ′ ′ = 2 k 2 \begin{align}
y_{p} &= k_2 x^2 + k_1 x + k_0 \\
y_{\rm p}' &= 2k_2 x + k_1 \\
y_{\rm p}'' &= 2k_2
\end{align} y p y p ′ y p ′′ = k 2 x 2 + k 1 x + k 0 = 2 k 2 x + k 1 = 2 k 2 So
y p ′ ′ + y p = 2 k 2 + k 2 x 2 + k 1 x + k 0 = 0.001 x 2 y_{\rm p}'' + y_{p} = 2 k_2 + k_2 x^2 + k_1 x + k_0 = 0.001x^2 y p ′′ + y p = 2 k 2 + k 2 x 2 + k 1 x + k 0 = 0.001 x 2 Equating coefficients of like powers of x gives the system
k 2 = 0.001 k 1 = 0 2 k 2 + k 0 = 0 \begin{align}
k_2 &= 0.001 \\
k_1 &= 0 \\
2k_2 + k_0 &= 0 \\
\end{align} k 2 k 1 2 k 2 + k 0 = 0.001 = 0 = 0 which gives k 0 = − 0.002 k_0 = -0.002 k 0 = − 0.002
Combine solutions and solve IVP.
The general solution and its first derivative are:
y = y h + y p = c 1 cos x + c 2 sin x + 0.001 x 2 − 0.002 y ′ = − c 1 sin x + c 2 cos x + 0.002 x \begin{align}
y &= y_{\rm h} + y_{\rm p} = c_1 \cos x + c_2 \sin x + 0.001x^2 - 0.002 \\
y' &= -c_1 \sin x + c_2 \cos x + 0.002x
\end{align} y y ′ = y h + y p = c 1 cos x + c 2 sin x + 0.001 x 2 − 0.002 = − c 1 sin x + c 2 cos x + 0.002 x Substituting in the first boundary condition:
y ( 0 ) = c 1 − 0.002 = 0 y ′ ( 0 ) = c 2 = 1.5 \begin{align}
y(0) &= c_1 - 0.002 = 0\\
y'(0) &= c_2 = 1.5
\end{align} y ( 0 ) y ′ ( 0 ) = c 1 − 0.002 = 0 = c 2 = 1.5 Hence, c 1 = 0.002 c_1 = 0.002 c 1 = 0.002 c 2 = 1.5 c_2 = 1.5 c 2 = 1.5
y = 0.002 cos x + 1.5 sin x + 0.001 x 2 − 0.002 y = 0.002 \cos x + 1.5 \sin x + 0.001x^2 - 0.002 y = 0.002 cos x + 1.5 sin x + 0.001 x 2 − 0.002 As a more complex example, let’s find the form of the particular solution for:
y ′ ′ + 3 y ′ + 2.25 y = − 10 e − 1.5 x + cos x y'' + 3y' + 2.25y = -10e^-1.5x + \cos x y ′′ + 3 y ′ + 2.25 y = − 10 e − 1.5 x + cos x Find y h y_{\rm h} y h
y h ′ ′ + 3 y h ′ + 2.25 y h = 0 λ 2 + 3 λ + 2.25 = 0 ( λ + 1.5 ) 2 = 0 \begin{align}
y_{\rm h}''+3y_{\rm h}'+2.25y_{\rm h} &= 0 \\
\lambda^2 +3\lambda+2.25 = 0 \\
(\lambda+1.5)^2 &= 0 \\
\end{align} y h ′′ + 3 y h ′ + 2.25 y h λ 2 + 3 λ + 2.25 = 0 ( λ + 1.5 ) 2 = 0 = 0 so λ = − 1.5 \lambda = -1.5 λ = − 1.5
y h = ( c 1 + c 2 x ) e − 1.5 x y_{\rm h} = (c_1 + c_2 x)e^{-1.5x} y h = ( c 1 + c 2 x ) e − 1.5 x Guess y p y_{\rm p} y p
We should add guesses for both terms in r . Our first term is an
exponential, but it needs to be multiplied by x 2 x^2 x 2
y p = k 1 x 2 e − 1.5 x + ( k 2 cos x + k 3 sin x ) \begin{align}
y_{\rm p} = k_1 x^2 e^{-1.5 x} + (k_2 \cos x +k_3 \sin x)
\end{align} y p = k 1 x 2 e − 1.5 x + ( k 2 cos x + k 3 sin x ) Skill builder problems ¶ Solve:
y ′ ′ + 3 y ′ + 2 y = 30 e 2 x , y ( 0 ) = 1 , y ′ ( 0 ) = 0 y'' + 3y' + 2y = 30e^{2x}, \quad y(0) = 1, \quad y'(0) = 0 y ′′ + 3 y ′ + 2 y = 30 e 2 x , y ( 0 ) = 1 , y ′ ( 0 ) = 0 Find homogeneous solution y h y_{\rm h} y h
y h ′ ′ + 3 y h ′ + 2 y h = 0 λ 2 + 3 λ + 2 = 0 ( λ + 1 ) ( λ + 2 ) = 0 \begin{align}
y_{\rm h}'' + 3y_{\rm h}' + 2y_{\rm h} &= 0 \\
\lambda^{2} + 3\lambda + 2 &= 0 \\
(\lambda + 1)(\lambda + 2) &= 0
\end{align} y h ′′ + 3 y h ′ + 2 y h λ 2 + 3 λ + 2 ( λ + 1 ) ( λ + 2 ) = 0 = 0 = 0 so λ 1 = − 1 \lambda_1 = -1 λ 1 = − 1 λ 2 = − 2 \lambda_2 = -2 λ 2 = − 2
y h = c 1 e x + c 2 e − 2 x y_{\rm h} = c_1 e^{x} + c_2 e^{-2x} y h = c 1 e x + c 2 e − 2 x Find particular solution y p y_{\rm p} y p
y p = k e 2 x y p ′ = 2 k e 2 x y p ′ ′ = 4 k e 2 x \begin{align}
y_{\rm p} &= ke^{2x} \\
y_{\rm p}' &= 2ke^{2x} \\
y_{\rm p}'' &= 4ke^{2x}
\end{align} y p y p ′ y p ′′ = k e 2 x = 2 k e 2 x = 4 k e 2 x Combine and solve for k :
4 k e 2 x e 2 x + 6 k e 2 x e 2 x + 2 k e 2 x e 2 x = 30 e 2 x e 2 x 12 k = 30 \begin{align}
4ke^{2x} e^{2x} &+ 6ke^{2x} e^{2x} + 2ke^{2x} e^{2x} = 30e^2x e^{2x} \\
12k &= 30
\end{align} 4 k e 2 x e 2 x 12 k + 6 k e 2 x e 2 x + 2 k e 2 x e 2 x = 30 e 2 x e 2 x = 30 so k = 5 / 2 k = 5/2 k = 5/2
y p = 5 2 e 2 x y_{\rm p} = \frac{5}{2} e^{2x} y p = 2 5 e 2 x Combine and apply initial conditions:
The general solution and its first derivative are
y = c 1 e − x + c 2 e − 2 x + 5 2 e 2 x y ′ = − c 1 e − x − 2 c 2 e − 2 x + 5 e 2 x \begin{align}
y &= c_1 e^{-x} + c_2 e^-{2x} + \frac{5}{2} e^{2x} \\
y' &= -c_1 e^{-x} - 2c_2 e^{-2x} + 5e^{2x}
\end{align} y y ′ = c 1 e − x + c 2 e − 2 x + 2 5 e 2 x = − c 1 e − x − 2 c 2 e − 2 x + 5 e 2 x Plugging in initial conditions
y ( 0 ) = c 1 + c 2 + 5 2 = 1 y ′ ( 0 ) = − c 1 − 2 c 2 + 5 = 0 \begin{align}
y(0) &= c_1 + c_2 + \frac{5}{2} = 1 \\
y'(0) &= -c_1 - 2c_2 + 5 = 0
\end{align} y ( 0 ) y ′ ( 0 ) = c 1 + c 2 + 2 5 = 1 = − c 1 − 2 c 2 + 5 = 0 This is a system of linear equations:
c 1 + c 2 = − 3 2 c 1 + 2 c 2 = 5 \begin{align}
c_1 + c_2 &= -\frac{3}{2} \\
c_1 + 2c_2 &= 5 \\
\end{align} c 1 + c 2 c 1 + 2 c 2 = − 2 3 = 5 that can be solved to give c 1 = − 8 c_1 = -8 c 1 = − 8 c 2 = 13 / 2 c_2 = 13/2 c 2 = 13/2
The final solution is:
y = − 8 e − x + 13 2 e − 2 x = 5 2 e 2 x y = -8e^{-x} + \frac{13}{2} e^{-2x} = \frac{5}{2} e^{2x} y = − 8 e − x + 2 13 e − 2 x = 2 5 e 2 x y ′ ′ + 4 y = 16 cos 2 x , y ( 0 ) , y ′ ( 0 ) = 0 y'' + 4y = 16 \cos 2x , \quad y(0), \quad y'(0) = 0 y ′′ + 4 y = 16 cos 2 x , y ( 0 ) , y ′ ( 0 ) = 0 Find homogeneous solution y h y_{\rm h} y h
y h ′ ′ + 4 y h = 0 λ 2 + 4 = 0 \begin{align}
y_{\rm h}'' + 4y_{\rm h} &= 0 \\
\lambda^{2} + 4 &= 0 \\
\end{align} y h ′′ + 4 y h λ 2 + 4 = 0 = 0 so λ 1 , 2 = ± 2 i \lambda_{1,2} = \pm 2i λ 1 , 2 = ± 2 i
y h = c 1 cos 2 x + c 2 sin 2 x y_{\rm h} = c_1 \cos 2x + c_2 \sin 2x y h = c 1 cos 2 x + c 2 sin 2 x Find the particular solution y p y_{\rm p} y p
y p = x ( k 1 cos 2 x + k 2 sin 2 x ) y p ′ = x ( − 2 k 1 sin 2 x + 2 k 2 cos 2 x ) + ( k 1 cos 2 x + k 2 sin 2 x ) y p ′ ′ = x ( − 4 k 1 cos 2 x − 4 k 2 sin 2 x ) + 2 ( − 2 k 1 sin 2 x + 2 k 2 cos 2 x ) \begin{align}
y_{\rm p} &= x(k_1 \cos 2x + k_2 \sin 2x) \\
y_{\rm p}' &= x(-2k_1 \sin 2x + 2k_2 \cos 2x) +
(k_1 \cos 2x + k_2 \sin 2x) \\
y_{\rm p}'' &= x(-4k_1 \cos 2x - 4k_2 \sin 2x) +
2(-2k_1 \sin 2x + 2k_2 \cos 2x)
\end{align} y p y p ′ y p ′′ = x ( k 1 cos 2 x + k 2 sin 2 x ) = x ( − 2 k 1 sin 2 x + 2 k 2 cos 2 x ) + ( k 1 cos 2 x + k 2 sin 2 x ) = x ( − 4 k 1 cos 2 x − 4 k 2 sin 2 x ) + 2 ( − 2 k 1 sin 2 x + 2 k 2 cos 2 x ) Combine:
y p ′ ′ + 4 y p = 2 ( − 2 k 1 sin 2 x + 2 k 2 cos 2 x ) = 16 cos 2 x y_{\rm p}'' + 4y_{\rm p} = 2(-2k_1 \sin 2x + 2k_2 \cos 2x) = 16\cos 2x y p ′′ + 4 y p = 2 ( − 2 k 1 sin 2 x + 2 k 2 cos 2 x ) = 16 cos 2 x Comparing terms gives k 1 = 0 k_1 = 0 k 1 = 0 k 2 = 4 k_2 = 4 k 2 = 4
Combine and apply initial conditions:
The general solution and its first derivative are
y = c 1 cos 2 x + c 2 sin 2 x + 4 x sin 2 x y ′ = − 2 c 1 sin 2 x + 2 c 2 cos 2 x + 4 x ( 2 cos 2 x ) + 4 sin 2 x \begin{align}
y &= c_1 \cos 2x + c_2 \sin 2x + 4x \sin 2x \\
y' &= -2c_1 \sin 2x + 2c_2 \cos 2x + 4x(2 \cos 2x) + 4 \sin 2x
\end{align} y y ′ = c 1 cos 2 x + c 2 sin 2 x + 4 x sin 2 x = − 2 c 1 sin 2 x + 2 c 2 cos 2 x + 4 x ( 2 cos 2 x ) + 4 sin 2 x Plugging in the initial conditions:
y ( 0 ) = c 1 = 0 y ′ ( 0 ) = 2 c 2 = 0 \begin{align}
y(0) &= c_1 = 0 \\
y'(0) &= 2c_2 = 0
\end{align} y ( 0 ) y ′ ( 0 ) = c 1 = 0 = 2 c 2 = 0 gives c 1 = 0 c_1 = 0 c 1 = 0 c 2 = 0 c_2 = 0 c 2 = 0
The final solution is:
y = 4 x sin 2 x y = 4x \sin 2x y = 4 x sin 2 x y ′ ′ − y ′ − 12 y = 144 x 3 + 25 2 , y ( 0 ) = 5 , y ′ ( 0 ) = − 1 / 2 y'' - y' - 12y = 144x^3 + \frac{25}{2}, \quad y(0) = 5, \quad y'(0) = -1/2 y ′′ − y ′ − 12 y = 144 x 3 + 2 25 , y ( 0 ) = 5 , y ′ ( 0 ) = − 1/2 Find homogeneous solution y h y_{\rm h} y h
y h ′ ′ − y h ′ − 12 y h = 0 λ 2 − λ − 12 = 0 ( λ − 4 ) ( λ + 3 ) = 0 \begin{align}
y_{\rm h}'' - y_{\rm h}' - 12y_{\rm h} &= 0
\lambda^2 - \lambda - 12 &= 0
(\lambda - 4)(\lambda + 3) &= 0
\end{align} y h ′′ − y h ′ − 12 y h = 0 λ 2 − λ − 12 = 0 ( λ − 4 ) ( λ + 3 ) = 0 so λ 1 = 4 \lambda_1 = 4 λ 1 = 4 λ 2 = − 3 \lambda_2 = -3 λ 2 = − 3
y h = c 1 e 4 x + c 2 e − 3 x y_{\rm h} = c_1 e^{4x} + c_2 e^{-3x} y h = c 1 e 4 x + c 2 e − 3 x Find the particular solution y p y_{\rm p} y p
y p = k 1 x 3 + k 2 x 2 + k 3 x + k 4 y p ′ = 3 k 1 x 2 + 2 k 2 x + k 3 y p ′ ′ = 6 k 1 x + 2 k 2 \begin{align}
y_{\rm p} &= k_1 x^3 + k_2 x^2 + k_3 x + k_4 \\
y_{\rm p}' &= 3k_1x^2 + 2k_2x + k_3 \\
y_{\rm p}'' &= 6k_1x + 2k_2
\end{align} y p y p ′ y p ′′ = k 1 x 3 + k 2 x 2 + k 3 x + k 4 = 3 k 1 x 2 + 2 k 2 x + k 3 = 6 k 1 x + 2 k 2 Substitute in ODE and combine:
y p ′ ′ − y p ′ − 12 y p = ( 6 k 1 x + 2 k 2 ) − ( 3 k 1 x 2 + 2 k 2 x + k 3 ) − 12 ( k 1 x 3 + k 2 x 2 + k 3 x + k 4 ) = − 12 k 1 x 3 + ( − 3 k 1 − 12 k 2 ) x 2 + ( 6 k 1 − 2 k 2 − 12 k 3 ) x + ( 2 k 2 − k 3 − 12 k 4 ) \begin{align}
& y_{\rm p}'' - y_{\rm p}' - 12y_{\rm p} \\
&= (6k_1x + 2k_2) - (3k_1x^2 + 2k_2x + k_3) -
12(k_1x^3 + k_2x^2 + k_3x + k_4) \\
&= -12k_1x^3 + (-3k_1 - 12k_2)x^2 + (6k_1 - 2k_2 - 12k_3)x +
(2k_2 - k_3 - 12k_4)
\end{align} y p ′′ − y p ′ − 12 y p = ( 6 k 1 x + 2 k 2 ) − ( 3 k 1 x 2 + 2 k 2 x + k 3 ) − 12 ( k 1 x 3 + k 2 x 2 + k 3 x + k 4 ) = − 12 k 1 x 3 + ( − 3 k 1 − 12 k 2 ) x 2 + ( 6 k 1 − 2 k 2 − 12 k 3 ) x + ( 2 k 2 − k 3 − 12 k 4 ) Comparing coefficients of powers of x gives:
− 12 k 1 = 144 − 3 k 1 − 12 k 2 = 0 36 − 12 k 2 = 0 6 k 1 − 2 k 2 − 12 k 3 = 0 2 k 2 − k 3 − 12 k 4 = 25 2 \begin{align}
-12k_1 &= 144 \\
-3k_1 - 12k_2 &= 0 \\
36 - 12k_2 &= 0 \\
6k_1 - 2k_2 - 12k_3 &= 0 \\
2k_2 - k_3 - 12k_4 &= \frac{25}{2}
\end{align} − 12 k 1 − 3 k 1 − 12 k 2 36 − 12 k 2 6 k 1 − 2 k 2 − 12 k 3 2 k 2 − k 3 − 12 k 4 = 144 = 0 = 0 = 0 = 2 25 Solving this system of linear equations gives k 1 = − 12 k_1 = -12 k 1 = − 12 k 2 = 3 k_2 = 3 k 2 = 3 k 3 = − 13 / 2 k_3 = -13/2 k 3 = − 13/2 k 4 = 0 k_4 = 0 k 4 = 0
y p = − 12 x 3 + 3 x 2 − 13 2 x y_{\rm p} = -12x_3 + 3x^2 - \frac{13}{2}x y p = − 12 x 3 + 3 x 2 − 2 13 x Combine and apply boundary conditions:
The general solution and its first derivative are
y = c 1 e 4 x + c 2 e − 3 x − 12 x 3 + 3 x 2 − 13 2 x y ′ = 4 c 1 e 4 x − 3 c 2 e − 3 X − 36 x 2 + 6 x − 13 2 \begin{align}
y &= c_1 e^{4x} + c_2 e^{-3x} - 12x^3 + 3x^2 - \frac{13}{2}x \\
y' &= 4c_1 e^{4x} - 3c_2e^{-3X} - 36x^2 + 6x - \frac{13}{2}
\end{align} y y ′ = c 1 e 4 x + c 2 e − 3 x − 12 x 3 + 3 x 2 − 2 13 x = 4 c 1 e 4 x − 3 c 2 e − 3 X − 36 x 2 + 6 x − 2 13 Plugging in the initial conditions:
y ( 0 ) = c 1 + c 2 = 5 y ′ ( 0 ) = 4 c 1 − 3 c 2 − 13 2 = − 1 2 \begin{align}
y(0) &= c_1 + c_2 = 5 \\
y'(0) &= 4c_1 - 3c_2 - \frac{13}{2} = -\frac{1}{2}
\end{align} y ( 0 ) y ′ ( 0 ) = c 1 + c 2 = 5 = 4 c 1 − 3 c 2 − 2 13 = − 2 1 Solve using matrices:
[ 1 1 5 4 − 3 6 ] R 1 − 4 R 1 → [ 1 1 5 0 − 7 14 ] R 1 ÷ − 7 → [ 1 1 5 0 1 2 ] − R 2 R 1 → [ 1 0 3 0 1 2 ] \begin{align}
\begin{bmatrix} 1 & 1 & 5 \\ 4 & -3 & 6 \end{bmatrix}
\begin{matrix} \vphantom{R_1} \\ -4 R_1 \end{matrix}
&\to \begin{bmatrix} 1 & 1 & 5 \\ 0 & -7 & 14 \end{bmatrix}
\begin{matrix} \vphantom{R_1} \\ \div -7 \end{matrix} \\
&\to \begin{bmatrix} 1 & 1 & 5 \\ 0 & 1 & 2 \end{bmatrix}
\begin{matrix} -R_2 \\ \vphantom{R_1} \end{matrix} \\
&\to \begin{bmatrix} 1 & 0 & 3 \\ 0 & 1 & 2 \end{bmatrix}
\end{align} [ 1 4 1 − 3 5 6 ] R 1 − 4 R 1 → [ 1 0 1 − 7 5 14 ] R 1 ÷ − 7 → [ 1 0 1 1 5 2 ] − R 2 R 1 → [ 1 0 0 1 3 2 ] so c 1 = 3 c_1 = 3 c 1 = 3 c 2 = 2 c_2 = 2 c 2 = 2
y = 3 e 4 x + 2 e − 3 x − 12 x 3 + 3 x 2 − 13 2 x y = 3e^{4x} + 2e^{-3x} - 12x^3 + 3x^2 - \frac{13}{2}x y = 3 e 4 x + 2 e − 3 x − 12 x 3 + 3 x 2 − 2 13 x