04 - Solutions to Inhomogeneous Linear Differential Equations

Finding the Solution to the Inhomogeneous Linear Differential Equation

The particular solution $y_p(x)$ is generally found by trial and error by mimicking the form of $f(x)$

Particular Solution similar to Complementary Function $x$

Multiply the previous particular solution by

Or when you solve the constant coefficients which then make $y_p(x)=0$

Example 1 - Particular Solution

Find the particular solution of

$$\frac{d^{2}y}{d{x}^2}-3\frac{dy}{dx}+2y=x$$

Solution

Try $y_p(x)=ax+b$ where $a,b\in \mathbb{R}$

So we get

$$\frac{d{y}_p}{dx}=a\text{ and }\frac{d^2{y}_p}{d{x}^2}=0$$

Substituting back we get

$$0-3a+2(ax+b)=x$$

As this holds for all $x$ then comparing coefficients

$$\begin{array}{rl}2a& =1⟹a=\frac{1}{2}\\ -3a+2b& =0⟹b=\frac{3}{4}\end{array}$$

Hence

$$y_p(x)=\frac{x}{2}+\frac{3}{4}$$

Example 2 - Particular Solution

Find the particular solution of

$$\frac{d^{2}y}{d{x}^2}+4\frac{dy}{dx}+4y=\sin (3x)$$

Solution

The auxiliary equation is $m^2+4m+4$ which has repeated root $-2$
Hence the complementary function is

$$y_c(x)=(C_{1}x+C_2)e^{-2x}$$

For the particular solution we will pick something like

$$y_p(x)=a\cos (3x)+b\sin (3x)$$

Differentiating and substituting gives you the equations

$$\begin{array}{rl}-9a+12b+4a& =0\\ -9b-12a+4b& =1\end{array}$$

Hence

$$a=-\frac{12}{169}\text{ and }b=-\frac{5}{169}$$

So we get particular solution

$$y_p(x)=-\frac{12}{169}\cos (3x)-\frac{5}{169}\sin (3x)$$

Example 3 - Particular Solution (Complementary Function Clash)

Fid the particular solution of

$$\frac{d^{2}y}{d{x}^2}-6\frac{dy}{dx}+9y=e^{3x},y(0)=y^{\prime }(0)=1$$

Solution $m^2-6y+9=0$ leads to the complementary function

The auxiliary equation

$$y_c(x)=(C_{1}x+C_2)e^{3x}$$

Our first guess for the particular solution is $a{e}^{3x}$
However as this is contained in $y_c(x)$ then we try $ax{e}^{3x}$
However again this again is also contained in $y_c(x)$

Hence our particular solution is in the form $a{x}^2{e}^{3x}$
Differentiating and substituting into our ODE gives that $\alpha =\frac{1}{2}$

Hence the particular solution is

$$y_p(x)=\frac{1}{2}{x}^2{e}^{3x}$$