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4.4 Undetermined Coefficients–Superposition Approach: Class Notes

4.4a

The techniques in this lesson apply to nonhomogeneous DEs of the form $a_n{y}^{(n)}+a_{n-1}{y}^{(n-1)}+\cdots +a_{1}y\prime +a_{0}y=g(x)$ whose input functions (that is, $g(x)$) are constants, polynomials, exponentials, sines and cosines, and sums and products of these.

From 4.1: Consider $y″-6y\prime +5y=-9e^{2x}+5x^2+3x-16$.

Here $g(x)=g_1(x)+g_2(x)$. The complete (general) solution is

$$y=c_1{e}^x+c_2{e}^{5x}+3e^{2x}+x^2+3x,$$

where $y_c=c_1{e}^x+c_2{e}^{5x}$ is the complementary function and $y_p=y_{p_1}+y_{p_2}$ is the particular solution. $y_c$ is a solution to the associated homogeneous DE $y″-6y\prime +5y=0$.

How do we find $y_p$ from scratch? Notice that $g(x)$ comprises an exponential function and a polynomial function.

Complementary function. Auxiliary equation:

$$m^2-6m+5=0\Rightarrow (m-1)(m-5)=0\Rightarrow m=1,5,y_c=c_1{e}^x+c_2{e}^{5x}.$$

First piece. Consider $y″-6y\prime +5y=-9e^{2x}$. Since $g_1(x)$ is exponential with exponent $2x$, guess

$$y_{p_1}=A{e}^{2x},y_{p_1}\prime =2A{e}^{2x},y_{p_1}″=4A{e}^{2x}.$$

Substitute:

$$\begin{array}{cc}\hfill {\displaystyle 4A{e}^{2x}-12A{e}^{2x}+5A{e}^{2x}}& {\displaystyle =-9e^{2x}}\hfill \\ \hfill {\displaystyle 4A-12A+5A=-3A}& {\displaystyle =-9\Rightarrow A=3}\hfill \end{array}$$

$$y_{p_1}=3e^{2x}$$

Second piece. Consider $y″-6y\prime +5y=5x^2+3x-16$. Since $g_2(x)$ is a 2nd-degree polynomial, guess a general 2nd-degree polynomial:

$$y_{p_2}=A{x}^2+Bx+C,y_{p_2}\prime =2Ax+B,y_{p_2}″=2A.$$

This goes into the equation:

$$\begin{array}{cc}\hfill {\displaystyle 2A-6(2Ax+B)+5(A{x}^2+Bx+C)}& {\displaystyle =5x^2+3x-16}\hfill \\ \hfill {\displaystyle (5A)x^2+(-12A+5B)x+(2A-6B+5C)}& {\displaystyle =5x^2+3x-16}\hfill \end{array}$$

Match coefficients, solving from the top down:

$$\begin{array}{cc}\hfill {\displaystyle x^2\text{ terms:}5A=5}& {\displaystyle \Rightarrow A=1}\hfill \\ \hfill {\displaystyle x\text{ terms:}-12A+5B=3}& {\displaystyle \Rightarrow B=3}\hfill \\ \hfill {\displaystyle \text{constants:}2A-6B+5C=-16}& {\displaystyle \Rightarrow C=0}\hfill \end{array}$$

$$y_{p_2}=x^2+3x$$

By superposition, $y_p=y_{p_1}+y_{p_2}=3e^{2x}+x^2+3x$, and the general solution to the nonhomogeneous DE is $y=y_c+y_p$:

Tip: We could have found both pieces at once with $y_p=A{e}^{2x}+B{x}^2+Cx+D$.

4.4b

Example

Solve the given differential equation by undetermined coefficients.

$$y″-8y\prime +20y=100x^2-26x{e}^x$$

Solution

Step 1: complementary function. Complete the square in the auxiliary equation:

$$\begin{array}{cc}\hfill {\displaystyle m^2-8m+20}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle m^2-8m+16}& {\displaystyle =-20+16}\hfill \\ \hfill {\displaystyle (m-4{)}^2}& {\displaystyle =-4\Rightarrow m=4\pm 2i}\hfill \end{array}$$

$$y_c=e^{4x}(c_1\cos 2x+c_2\sin 2x)$$

Step 2: the polynomial piece $100x^2$.

$$y_{p_1}=A{x}^2+Bx+C,y_{p_1}\prime =2Ax+B,y_{p_1}″=2A$$

$$\begin{array}{cc}\hfill {\displaystyle 2A-8(2Ax+B)+20(A{x}^2+Bx+C)}& {\displaystyle =100x^2}\hfill \\ \hfill {\displaystyle 20A{x}^2+(-16A+20B)x+(2A-8B+20C)}& {\displaystyle =100x^2}\hfill \end{array}$$

Matching coefficients: $20A=100$, $-16A+20B=0$, and $2A-8B+20C=0$, so

$${\displaystyle A=5,B=4,C=\frac{11}{10},y_{p_1}=5x^2+4x+\frac{11}{10}.}$$

Step 3: the piece $-26x{e}^x$. It is a product of a linear function and an exponential, so

$$y_{p_2}=(Ax+B)e^x.$$

By the product rule (factoring out $e^x$),

$$\begin{array}{cc}\hfill {\displaystyle y_{p_2}\prime }& {\displaystyle =A{e}^x+(Ax+B)e^x=(A+Ax+B)e^x}\hfill \\ \hfill {\displaystyle y_{p_2}″}& {\displaystyle =(A+A+Ax+B)e^x=(2A+B+Ax)e^x.}\hfill \end{array}$$

Substitute into $y″-8y\prime +20y=-26x{e}^x$ and cancel $e^x$:

$$(2A+B+Ax)-8(A+B+Ax)+20(Ax+B)=-26x$$

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle x\text{ terms:}A-8A+20A=-26}& {\displaystyle \Rightarrow A=-2}\hfill \\ \hfill {\displaystyle \text{constants:}2A+B-8A-8B+20B=0}& {\displaystyle \Rightarrow B=-\frac{12}{13}}\hfill \end{array}}$$

So ${\displaystyle y_{p_2}=(-2x-\frac{12}{13})e^x}$, and $y=y_c+y_{p_1}+y_{p_2}$:

4.4c

Example

$y″-16y=2e^{4x}$

Solution

Setting ourselves up for potential failure: since $g(x)=2e^{4x}$, try

$$y_p=A{e}^{4x},y_p\prime =4A{e}^{4x},y_p″=16A{e}^{4x}.$$

$$16A{e}^{4x}-16A{e}^{4x}=2e^{4x}\Rightarrow 0=2e^{4x}\text{(impossible)}$$

We must always first find $y_c$:

$$m^2-16=0\Rightarrow m=\pm 4,y_c=c_1{e}^{4x}+c_2{e}^{-4x}.$$

The guess $y_p=A{e}^{4x}$ is redundant with $y_c$: it is a solution of the homogeneous equation, so it produces 0 on the left side. To find a form of $y_p$ that is independent with $y_c$, we insert factor(s) of $x$. Here

$$y_p=Ax{e}^{4x},$$

where this factor of $x$ avoids redundancy with $y_c$. Then

$$\begin{array}{cc}\hfill {\displaystyle y_p\prime }& {\displaystyle =(A+4Ax)e^{4x}}\hfill \\ \hfill {\displaystyle y_p″}& {\displaystyle =(4A+4A+16Ax)e^{4x}=(8A+16Ax)e^{4x}.}\hfill \end{array}$$

Substitute into $y″-16y=2e^{4x}$ and cancel $e^{4x}$:

$${\displaystyle 8A+16Ax-16Ax=2\Rightarrow A=\frac{1}{4}.}$$

Recall $y=y_c+y_p$:

Example

$4y″-4y\prime -3y=\cos 2x$

Solution

$${\displaystyle 4m^2-4m-3=0\Rightarrow (2m+1)(2m-3)=0\Rightarrow m=-\frac{1}{2},\frac{3}{2}}$$

$$y_c=c_1{e}^{-x/2}+c_2{e}^{3x/2}$$

For $g(x)=\cos 2x$, the guess needs both cosine and sine:

$$\begin{array}{cc}\hfill {\displaystyle y_p}& {\displaystyle =A\cos 2x+B\sin 2x}\hfill \\ \hfill {\displaystyle y_p\prime }& {\displaystyle =-2A\sin 2x+2B\cos 2x}\hfill \\ \hfill {\displaystyle y_p″}& {\displaystyle =-4A\cos 2x-4B\sin 2x}\hfill \end{array}$$

Multiply by the coefficients $4$, $-4$, $-3$ and substitute:

$$\begin{array}{cc}\hfill {\displaystyle -16A\cos 2x-16B\sin 2x+8A\sin 2x-8B\cos 2x}& {\displaystyle }\hfill \\ \hfill {\displaystyle -3A\cos 2x-3B\sin 2x}& {\displaystyle =\cos 2x}\hfill \end{array}$$

$$\begin{array}{cc}\hfill {\displaystyle \text{Cosine terms:}-16A-8B-3A}& {\displaystyle =1\Rightarrow -19A-8B=1}\hfill \\ \hfill {\displaystyle \text{Sine terms:}-16B+8A-3B}& {\displaystyle =0\Rightarrow 8A-19B=0}\hfill \end{array}$$

Solve by elimination. Multiply the first equation by $-19$ and the second by $8$, then add:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle 361A+8(19)B}& {\displaystyle =-19}\hfill \\ \hfill {\displaystyle 64A-8(19)B}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle 425A}& {\displaystyle =-1\Rightarrow A=-\frac{19}{425}}\hfill \end{array}}$$

Multiply the first equation by $8$ and the second by $19$, then add:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle -8(19)A-64B}& {\displaystyle =8}\hfill \\ \hfill {\displaystyle 8(19)A-361B}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle -425B}& {\displaystyle =8\Rightarrow B=-\frac{8}{425}}\hfill \end{array}}$$

$y_p$ forms, assuming no redundancy