Question

In Problems 27-36solve the given initial-value problem.

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathit{y}\mathbf{=}\mathit{c}\mathit{o}\mathit{s}\mathit{h}\mathit{x}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{,}\mathit{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{12}$

Short answer

$y=-5e^{-x}+7e^x-\frac{1}{4}x{e}^{-x}+\frac{1}{4}x{e}^x$

Step-by-step solution

Complementary function

Consider the differential equation$y\text{'}\text{'}-y=\cosh x,y\left(0\right)=2,y\text{'}\left(0\right)=12$

Rewrite the differential equation as,

$y\text{'}\text{'}-y=\frac{e^x+e^{-x}}{2}.$

The auxiliary equation corresponding to homogenous equation is:$m^2-1=0.$

$$\begin{gathered} m^2-1=0 \\ m=\pm 1 \end{gathered}$$

Therefore, the complementary function is$y_c=c_1{e}^{-x}+c_2{e}^x$

For particular solution

Let particular solution be.$y_p=Ax{e}^{-x}+Bx{e}^x$

Differentiating we have:

$$\begin{gathered} y_p\text{'}=A{e}^{-x}-Ax{e}^{-x}+B{e}^x+Bx{e}^x \\ {y}_p\text{'}\text{'}=-A{e}^{-x}-A{e}^{-x}+Ax{e}^{-x}+B{e}^x+B{e}^x+Bx{e}^x \\ =-2A{e}^{-x}+Ax{e}^{-x}+2B{e}^x+Bx{e}^x \end{gathered}$$

Substitute these values in.

$y_p\text{'}\text{'}-y_p=\frac{e^x+e^{-x}}{2}$

$$\begin{gathered} y_p\text{'}\text{'}-y_p=\frac{e^x+e^{-x}}{2} \\ -2A{e}^{-x}+Ax{e}^{-x}+2B{e}^x+Bx{e}^x-Ax{e}^{-x}-Bx{e}^x=\frac{e^x+e^{-x}}{2} \\ -2A{e}^{-x}+2B{e}^x=\frac{1}{2}{e}^x+\frac{1}{2}{e}^{-x} \end{gathered}$$

Equate the coefficients of like terms on both sides.

$-2A=\frac{1}{2} \text{and} 2B=\frac{1}{2}.$

$A=-\frac{1}{4}$and$B=\frac{1}{4}$

Therefore, the particular solution is.

$y_p=-\frac{1}{4}x{e}^{-x}+\frac{1}{4}x{e}^x$

General solution

The general solution of the IVP can be written as,

$$\begin{gathered} y=y_c+y_p \\ =c_1{e}^{-x}+c_2{e}^x-\frac{1}{4}x{e}^{-x}+\frac{1}{4}x{e}^x \end{gathered}$$

Therefore, the solution of the IVP is$y=c_1{e}^{-x}+c_2{e}^x-\frac{1}{4}x{e}^{-x}+\frac{1}{4}x{e}^x$

Apply initial condition

Put$y\left(0\right)=2$in$y=c_1{e}^{-x}+c_2{e}^x-\frac{1}{4}x{e}^{-x}+\frac{1}{4}x{e}^x$

$$\begin{gathered} y\left(0\right)=c_1{e}^0+c_2{e}^0-\frac{1}{4}\left(0\right)e^0+\frac{1}{4}\left(0\right)e^0 \\ 2=c_1+c_2 \end{gathered}$$

Differentiate with respect to x.

$y\text{'}=-c_1{e}^{-x}+c_2{e}^x-\frac{1}{4}{e}^{-x}+\frac{1}{4}x{e}^{-x}+\frac{1}{4}{e}^x+\frac{1}{4}x{e}^x.$

Put$y\text{'}\left(0\right)=12$in.$y\text{'}=-c_1{e}^{-x}+c_2{e}^x-\frac{1}{4}x{e}^{-x}+\frac{1}{4}x{e}^{-x}+\frac{1}{4}{e}^x+\frac{1}{4}x{e}^x$

$y\text{'}\left(0\right)=-c_1{e}^0+c_2{e}^0-\frac{1}{4}{e}^0+\frac{1}{4}\left(0\right)e^0+\frac{1}{4}{e}^0+\frac{1}{4}\left(0\right)e^0$

localid="1664111077012" $$\begin{gathered} 12=-c_1+c_2-\frac{1}{4}+\frac{1}{4} \\ 12=-c_1+c_2 \end{gathered}$$

Solve the equations

Then $c_1=-5$and $c_2=7$

Therefore, the solution is

$\mathit{y}\mathbf{=}\mathbf{-}\mathbf{5}{\mathit{e}}^{\mathbf{-}\mathbf{x}}\mathbf{+}\mathbf{7}{\mathit{e}}^{\mathbf{x}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{4}}\mathit{x}{\mathit{e}}^{\mathbf{-}\mathbf{x}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{4}}\mathit{x}{\mathit{e}}^{\mathbf{x}}$