Question

verify that the given two-parameter family of functions is the general solution of the nonhomogeneous differential equation on the indicated interval.

(a)Verify that ${\mathit{y}}_{\mathbf{p}\mathbf{1}}\mathbf{=}\mathbf{3}{\mathit{e}}^{\mathbf{2}\mathbf{x}}$and ${\mathit{y}}_{\mathbf{p}\mathbf{2}}\mathbf{=}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathbf{3}\mathit{x}$ are respectively, particular solutions of $\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{6}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathit{y}\mathbf{=}\mathbf{-}\mathbf{9}{\mathit{e}}^{\mathbf{2}\mathbf{x}}$ and $\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{6}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathit{y}\mathbf{=}\mathbf{5}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathbf{3}\mathit{x}\mathbf{-}\mathbf{16}$

(b) Use part (a) to find particular solutions of $\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{6}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathit{y}\mathbf{=}\mathbf{5}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathbf{3}\mathit{x}\mathbf{-}\mathbf{16}\mathbf{-}\mathbf{9}{\mathit{e}}^{\mathbf{2}\mathbf{x}}$and$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{6}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathit{y}\mathbf{=}\mathbf{-}\mathbf{10}{\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{6}\mathit{x}\mathbf{+}\mathbf{32}\mathbf{+}{\mathit{e}}^{\mathbf{2}\mathbf{x}}$

Short answer

(a) Proved

(b)$y=3e^{2x}+x^2+3xandy=-\frac{1}{3}{e}^{2x}-2x^2-6x$$y=3e^{2x}+x^2+3xandy=-\frac{1}{3}{e}^{2x}-2x^2-6x$

Step-by-step solution

Step 1(a): check the homogenous of function

We need to verify that the given functions $y_p$ are particular solutions of the given non homogeneous differential equations by showing that the function satisfies the equation. For the function$y_{p_1}\left(x\right)=3e^{2x}$ , we have

$y\text{'}=6e^{2x}andy\text{'}\text{'}=12e^{2x}$

Substituting the above derivatives on the left hand side of the first non-homogenous DE gives,

$\begin{array}{rcl}y\text{'}\text{'}-6y\text{'}+5y& =& 12e^{2x}-6\left(6e^{2x}\right)+5\left(3e^{2x}\right)\\ & =& 12e^{2x}-36e^{2x}+15e^{2x}\\ & =& -9e^{2x}\end{array}$

which is same as the right hand side of the nonhomogeneous differential equation. Hence, the function$y_{p_1}=3e^{2x}$ is the particular solution of the non-homogeneous differential equation

$y\text{'}\text{'}-6y\text{'}+5y=-9e^{2x}······\left(1\right)$

For the function$y_{p2}\left(x\right)=x^2+3x$ , we have

$y\text{'}=2x+3andy\text{'}\text{'}=2$

Substituting the above derivatives on the left hand side of the second non-homogenous DE gives,

$\begin{array}{rcl}y\text{'}\text{'}-6y\text{'}+5y& =& 2-6(2x+3)+5\left(x^2+3x\right)\\ & =& 2-12x-18+5x^2+15x\\ & =& 5x^2+3x-16\end{array}$

which is same as the right hand side of the nonhomogeneous differential equation. Hence, the function$y_{p2}\left(x\right)=x^2+3x$is the particular solution of the non-homogeneous differential equation$y\text{'}\text{'}-6y\text{'}+5y=5x^2+3x-16$

Hence it is verified that $y_{p1}=3e^{2x}$ and $y_{p2}=x^2+3x$ are respectively, particular solutions of $y\text{'}\text{'}-6y\text{'}+5y=-9e^{2x}$ and $y\text{'}\text{'}-6y\text{'}+5y=5x^2+3x-16$

Step 2(b): Find the y value

From part (a), we saw that$y_{p_1}=3e^{2x}$is the particular solution of

$y\text{'}\text{'}-6y\text{'}+5y=-9e^{2x}$

And $y_{p2}\left(x\right)=x^2+3x$ is the particular solution of

$y\text{'}\text{'}-6y\text{'}+5y=5x^2+\underset{g_2\left(x\right)}{3}x-16+-9{\underset{g_1\left(x\right)}{e}}^{2x}$

Is the superposition of $y_{p1}and{y}_{p2}$ie.,

$\begin{array}{rcl}y& =& y_{p_1}+y_{p_2}\\ & =& 3e^{2x}+x^2+3x\end{array}$

Verify the solution

Similarly, the second equation can be written as

$$\begin{gathered} y\text{'}\text{'}-6y\text{'}+5y=-10x^2-6x+32+e^{2x} \\ =-2\left(5x^2+3x-16\right)+\left(-\frac{1}{9}\right)\left(-9e^{2x}\right) \\ =-2(5x^2+\underset{g_2\left(x\right)}{3}x-16)+\left(-\frac{1}{9}\right)\left(-9{\underset{g_1\left(x\right)}{e}}^{2x}\right) \end{gathered}$$

From part (a), we saw that $y_{p_1}=3e^{2x}$is the particular solution of

$y\text{'}\text{'}-6y\text{'}+5y=-9e^{2x}$

And $y_{p2}\left(x\right)=x^2+3x$ is the particular solution of

$y\text{'}\text{'}-6y\text{'}+5y=5x^2+3x-16$

$y\text{'}\text{'}-6y\text{'}+5y=-2(5x^2+\underset{g_2\left(x\right)}{3}x-16)+\left(-\frac{1}{9}\right)\left(-9{\underset{g_1\left(x\right)}{e}}^{2x}\right)$

Is the superposition of $-2y_{p_2}and-\frac{1}{9}{y}_{p_1}\text{,}$ie.,

$\begin{array}{rcl}y& =& -\frac{1}{9}{y}_{p_1}-2y_{p_2}\\ & =& \left(-\frac{1}{9}\right)\left(3e^{2x}\right)-2\left(x^2+3x\right)\\ & =& -\frac{1}{3}{e}^{2x}-2x^2-6x\end{array}$

Hence the particular solutions are $y=3e^{2x}+x^2+3xandy=-\frac{1}{3}{e}^{2x}-2x^2-6x$