# Undetermined Coefficients

by Justin Skycak on

Undetermined coefficients can help us find a solution to a linear differential equation with constant coefficients when the right hand side is not equal to zero.

This post is part of the book Justin Math: Calculus. Suggested citation: Skycak, J. (2019). Undetermined Coefficients. In Justin Math: Calculus. https://justinmath.com/undetermined-coefficients/

In the previous post, we learned how to solve differential equations of the form

\begin{align*} a_ny^{(n)} + \cdots + a_2 y'' + a_1 y' + a_0 y = 0. \end{align*}

Now, we consider differential equations of the form

\begin{align*} a_ny^{(n)} + \cdots + a_2 y'' + a_1 y' + a_0 y = f(x) \end{align*}

where the right hand side is no longer strictly $0$, but rather some function $f(x)$. The solution to such a differential equation is given by

\begin{align*} y(x) = y_0(x) + y_f(x) \end{align*}

where $y_0$ is the general solution to the “homogeneous” equation

\begin{align*} a_ny^{(n)} + \cdots + a_2 y'' + a_1 y' + a_0 y = 0 \end{align*}

and $y_f$ is a particular solution that satisfies the “inhomogeneous” equation

\begin{align*} a_ny^{(n)} + \cdots + a_2 y'' + a_1 y' + a_0 y = f(x). \end{align*}

We already know how to use the characteristic polynomial to find $y_0$, and now we will learn how to use the method of undetermined coefficients to find $y_f$.

The method of undetermined coefficients involves guessing a solution $y_f(x)$ having the same form as $f(x)$, except possibly multiplied by some other coefficients. We then substitute this guess into the differential equation, and solve for the value of the coefficient that will make the guess correct.

## Case of Exponential Function

For example, to find a particular solution to the differential equation $y’^\prime+2y’+y=1-2e^{3x}$, we can guess that $y_f(x)=A+Be^{3x}$ for some values of $A$ and $B$. Substituting our guess into the equation, we can solve for the correct values of $A$ and $B$.

\begin{align*} y''+2y'+y &= 1-2e^{3x} \\ \left( A+Be^{3x} \right)'' + 2 \left( A+Be^{3x} \right)' + A+Be^{3x} &= 1-2e^{3x} \\ 9Be^{3x} + 6Be^{3x} + A+Be^{3x} &= 1-2e^{3x} \\ 16Be^{3x} + A &= 1-2e^{3x} \\ A = 1, & \hspace{.25cm} B = -\frac{1}{8} \end{align*}

Our particular solution is then given by $y_f(x)=1-\frac{1}{8}e^{3x}$. Then, using the characteristic polynomial method, we solve $y’^\prime+2y’+y=0$ to find $y_0(x)=(C_1+C_2)e^{-x}$. The full solution to the differential equation $y’^\prime+2y’+y=1-2e^{3x}$ is then given by

\begin{align*} y(x) &= y_0(x) + y_f(x) \\ &= (C_1+C_2x)e^{-x} + 1 - \frac{1}{8}e^{3x}. \end{align*}

## Case of Trig Functions with Same Angle

In cases where $f(x)$ contains $\sin \theta$ or $\cos \theta$, we include both $\sin \theta$ and $\cos \theta$ in our guess for $y_0$.

For example, to find a particular solution to the differential equation $y’’^\prime-3y’^\prime+2y’=-3\cos 2x$, we need to construct a guess that contains both $\sin 2x$ and $\cos 2x$. Our guess, then, is

\begin{align*} y_f(x) = A \sin 2x + B \cos 2x. \end{align*}

We substitute this guess into the differential equation and simplify.

\begin{align*} \begin{matrix} \left( A \sin 2x + B \cos 2x \right)''' - 3\left( A \sin 2x + B \cos 2x \right)'' \\ + 2\left( A \sin 2x + B \cos 2x \right)' \end{matrix} &= -3\cos 2x \\ \begin{matrix} \text{ } \\ \left( -8A \cos 2x + 8B \sin 2x \right) - 3\left( -4A \sin 2x -4B \cos 2x \right) \\ + 2\left( 2A \cos 2x - 2B \sin 2x \right) \\ \text{ } \end{matrix} &= -3\cos 2x \\ (12A+4B) \sin 2x + (-4A+12B) \cos 2x &= -3\cos 2x \end{align*}

Equating coefficients on the left and right sides of the equation yields a system of equations for $A$ and $B$.

\begin{align*} \begin{cases} 12A+4B &= 0 \\ -4A+12B &= -3 \end{cases} \end{align*}

Solving this system, we find $A=\frac{3}{40}$ and $B=-\frac{9}{40}$. The particular solution is then $y_f(x) = \frac{3}{40} \sin 2x - \frac{9}{40} \cos 2x$.

Using the characteristic polynomial to solve $y’’^\prime-3y’^\prime+2y’=0$ yields $y_0(x)=C_1+C_2 e^x + C_3 e^{2x}$, and the full solution of the differential equation $y’’^\prime-3y’^\prime+2y’=-3\cos 2x$ is then given by

\begin{align*} y(x) &= y_0(x) + y_f(x) \\ &= C_1 + C_2 e^x + C_3 e^{2x} + \frac{3}{40} \sin 2x - \frac{9}{40} \cos 2x. \end{align*}

## Case of Trig Functions with Different Angles

When we have multiple values of $\theta$, we end up with even more unknown coefficients in our guess.

For example, to find a particular solution to the differential equation $y’^\prime-2y=4\sin 3x - \cos 5x$, we need to construct a guess that contains both $\sin \theta$ and $\cos \theta$, for both $\theta = 3x$ and $\theta = 5x$. Our guess, then, is

\begin{align*} y_f(x) = A\sin 3x + B\cos 3x + C \sin 5x + D \cos 5x. \end{align*}

We substitute this guess into the differential equation and simplify.

\begin{align*} \begin{matrix} \left( A\sin 3x + B\cos 3x + C \sin 5x + D \cos 5x \right)'' \\ - 2\left( A\sin 3x + B\cos 3x + C \sin 5x + D \cos 5x \right) \end{matrix} &= 4\sin 3x - \cos 5x \\ \begin{matrix} \text{ } \\ \left( -9A\sin 3x -9 B \cos 3x - 25C \sin 5x - 25D \cos 5x \right) \\ - 2\left( A\sin 3x + B\cos 3x + C \sin 5x + D \cos 5x \right) \\ \text{ } \end{matrix} &= 4\sin 3x - \cos 5x \\ -11A \sin 3x - 11B \cos 3x - 27C \sin 5x - 27D \cos 5x &= 4\sin 3x - \cos 5x \end{align*}

Equating coefficients on the left and right sides of the equation yields $A=-\frac{4}{11}$, $B=0$, $C=0$, and $D=\frac{1}{27}$. The particular solution is then $y_f(x) = -\frac{4}{11} \sin 3x + \frac{1}{27} \cos 5x$.

Using the characteristic polynomial to solve $y’^\prime-2y=0$ yields $y_0(x) = C_1e^{\sqrt{2} x} + C_2e^{-\sqrt{2} x}$, and the full solution of the differential equation $y’^\prime-2y=4\sin 3x -\cos 5x$ is then given by

\begin{align*} y(x) &= y_0(x) + y_f(x) \\ &= C_1e^{\sqrt{2} x} + C_2e^{-\sqrt{2} x} -\frac{4}{11} \sin 3x + \frac{1}{27} \cos 5x . \end{align*}

## Case of Polynomial Functions

Lastly, the differential equation $y’^\prime+y’=x^3-2x+e^{2x}$ has a polynomial and an exponential term, so our guess for the particular solution needs to contain a polynomial and an exponential term.

The polynomial in the differential equation is of degree $3$, and the differential equation itself is of degree $2$, so our guess needs to contain a polynomial of degree $3+2=5$.

\begin{align*} y_f(x) = Ax^5 + Bx^4 + Cx^3 + Dx^2 + Ex + F + Ge^{2x} \end{align*}

We substitute this guess into the differential equation and simplify.

\begin{align*} \begin{matrix} \left( Ax^5 + Bx^4 + Cx^3 + Dx^2 + Ex + F + Ge^{2x} \right)'' \\ + \left( Ax^5 + Bx^4 + Cx^3 + Dx^2 + Ex + F + Ge^{2x} \right)' \end{matrix} &= x^3-2x+e^{2x} \\ \begin{matrix} \text{ } \\ \left( 20Ax^3 + 12Bx^2 + 6Cx + 2D + 4Ge^{2x} \right) \\ + \left( 5Ax^4 + 4Bx^3 + 3Cx^2 + 2Dx + E + 2Ge^{2x} \right) \\ \text{ } \end{matrix} &= x^3-2x+e^{2x} \\ \begin{matrix} 5Ax^4+(20A+4B)x^3+(12B+3C)x^2 \\ + (6C+2D)x+2D+E + 6Ge^{2x} \end{matrix} &= x^3-2x+e^{2x} \end{align*}

Equating coefficients on the left and right sides of the equation yields $A=0$, $B=\frac{1}{4}$, $C=-1$, $D=2$, $E=-4$, and $G=\frac{1}{6}$. The coefficient $F$ can still be any number, so we leave it as-is. The particular solution is then

\begin{align*} y_f(x) = \frac{1}{4}x^4 -x^3 +2x^2 -4x + F +\frac{1}{6} e^{2x} . \end{align*}

Using the characteristic polynomial to solve $y’^\prime+y’=0$ yields $y_0(x) = C_1 + C_2 e^has {-x}$, and the full solution of the differential equation $y’^\prime+y’=x^3-2x+e^{2x}$ is then given by

\begin{align*} y(x) &= y_0(x) + y_f(x) \\ &= C_1 + C_2e^{-x} +\frac{1}{4}x^4 -x^3 +2x^2 -4x + F +\frac{1}{6} e^{2x}. \end{align*}

To eliminate redundancy, we can lump the $F$ constant into the $C_1$ constant, since $C_1+F$ is itself just another constant.

\begin{align*} y(x) = C_1 + C_2e^{-x} +\frac{1}{4}x^4 -x^3 +2x^2 -4x +\frac{1}{6} e^{2x} \end{align*}

## Exercises

Use the method of undetermined coefficients to solve the following differential equations. (You can view the solution by clicking on the problem.)

\begin{align*} 1) \hspace{.5cm} y''+y=4e^{5x} \end{align*}
Solution:
\begin{align*} y=C_1 \cos x + C_2 \sin x + \frac{2}{13}e^{5x} \end{align*}

\begin{align*} 2) \hspace{.5cm} y'+3y=\sin(2x) \end{align*}
Solution:
\begin{align*} y=C_1e^{-3x}+\frac{3\sin 2x - 2\cos 2x}{13} \end{align*}

\begin{align*} 3) \hspace{.5cm} y''-y'=\cos(\pi x) \end{align*}
Solution:
\begin{align*} y=C_1e^x + C_2 - \frac{\sin (\pi x)+ \pi \cos (\pi x)}{\pi + \pi^3} \end{align*}

\begin{align*} 4) \hspace{.5cm} y'''-2y'=x^2+1 \end{align*}
Solution:
\begin{align*} y=C_1e^{\sqrt{2}x} + C_2e^{-\sqrt{2}x} + C_3 - \frac{1}{6}x^3-x \end{align*}

\begin{align*} 5) \hspace{.5cm} 2y'-y=\sin(x)-\cos(2x) \end{align*}
Solution:
\begin{align*} y=C_1 e^{\frac{1}{2}x}-\frac{\sin x + 2 \cos x}{5} + \frac{\cos(2x)-4\sin(2x)}{17} \end{align*}

\begin{align*} 6) \hspace{.5cm} 2y'+y=e^x+3\sin(x) \end{align*}
Solution:
\begin{align*} y=C_1e^{-\frac{1}{2}x} + \frac{1}{3}e^{x} + \frac{3\sin x- 6\cos x}{5} \end{align*}

\begin{align*} 7) \hspace{.5cm} 4y''-9y=2x^4-3x^2+\cos(x+1) \end{align*}
Solution:
\begin{align*} y=C_1e^{\frac{3}{2}x} + C_2e^{-\frac{3}{2}x} - \frac{2}{9}x^4-\frac{23}{27}x^2-\frac{1}{13}\cos(x+1)-\frac{184}{243} \end{align*}

\begin{align*} 8) \hspace{.5cm} y'+y=\sin(2x+1)+\cos(5x)+1 \end{align*}
Solution:
\begin{align*} y=C_1e^{-x} + \frac{5\sin(5x)+\cos(5x)}{26} + \frac{\sin(2x+1)-2\cos(2x+1)}{5}+1 \end{align*}

This post is part of the book Justin Math: Calculus. Suggested citation: Skycak, J. (2019). Undetermined Coefficients. In Justin Math: Calculus. https://justinmath.com/undetermined-coefficients/

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