Question

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

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{2}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathit{y}\mathbf{\text{'}}\mathbf{=}\mathbf{2}\mathbf{-}\mathbf{24}{\mathit{e}}^{\mathbf{x}}\mathbf{+}\mathbf{40}{\mathit{e}}^{\mathbf{5}\mathbf{x}}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{,}\mathit{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\frac{\mathbf{5}}{\mathbf{2}}\mathbf{,}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\frac{\mathbf{9}}{\mathbf{2}}$

Short answer

$y\left(x\right)=11-11e^x+9x{e}^x+2x-12x^2{e}^x+\frac{1}{2}{e}^{5x}$

Step-by-step solution

Complementary function

Consider the differential equation

$y^{\text{'}\text{'}\text{'}}-2y^{\text{'}\text{'}}+y^{\text{'}}=2-24e^x+40e^{5x},y\left(0\right)=\frac{1}{2},y^{\text{'}}\left(0\right)=\frac{5}{2},y^{\text{'}\text{'}}\left(0\right)=-\frac{9}{2} ······\left(1\right)$

The corresponding homogeneous differential equation is, $y^{\text{'}\text{'}\text{'}}-2y^{\text{'}\text{'}}+y^{\text{'}}=0$

The auxiliary equation of the associated homogeneous differential equation is

$$\begin{gathered} m^3-2m^2+m=0 \\ m\left(m^2-2m+1\right)=0 \\ m(m-1{)}^2=0 \\ m=0,1,1 \end{gathered}$$

Therefore, the complementary solution is

$y_c=c_1+c_2{e}^x+c_{3}x{e}^x······\left(2\right)$

Particular solution

Let us assume the particular solution is of the form $y_p=Ax+B{x}^2{e}^x+C{e}^{5x}$and it satisfies the equation (1).

That is,$y_p^{\text{'}\text{'}\text{'}}-2y_p^{\text{'}\text{'}\text{'}}+y_p^{\text{'}}=2-24e^x+40e^{5x}······\left(3\right)$

Now find the derivatives ofas shown below:

$$\begin{gathered} y_p^{\text{'}}=A+B{e}^x\left(x^2+2x\right)+5C{e}^{5x} \\ {y}_p^{\text{'}\text{'}}=B\left[x^2{e}^x+4x{e}^x+2e^x\right]+25C{e}^{5x} \\ {y}_p^{\text{'}\text{'}\text{'}}=B{e}^x\left[x^2+6x+6\right]+125C{e}^{5x} \end{gathered}$$

Substitute the values of $y_p^{\text{'}},y^{\text{'}\text{'}}$and $y_p^{\text{'}\text{'}\text{'}}$values is (3), then

$$\begin{gathered} B\left(x^2+6x+6\right)e^x+125C{e}^{5x}-2B\left(x^2+4x+2\right)e^x-50C{e}^{5x} \\ +A+B\left(x^2+2x\right)e^x+5C{e}^{5x}=2-24e^x+40e^{5x} \\ \left(x^2+6x+6-2x^2-8x-4+x^2+2x\right)B{e}^x+(125-50+5)C{e}^{5x}+A=2-24e^x+40e^{5x} \\ 2B{e}^x+80C{e}^{5x}+A=2-24e^x+40e^{5x} \end{gathered}$$

So, particular solution became:

$y_p=2x-12x^2{e}^x+\frac{1}{2}{e}^{5x}$

Differentiate

Therefore, the general solution is

$$\begin{gathered} y\left(x\right)=y_c+y_p \\ y\left(x\right)=c_1+c_2{e}^x+c_{3}x{e}^x+2x-12x^2{e}^x+\frac{1}{2}{e}^{5x} \end{gathered}$$

Differentiate the solution:

$$\begin{gathered} y^{\text{'}}\left(x\right)=c_2{e}^x+c_3\left(x{e}^x+e^x\right)+2-12\left(x^2{e}^x+2x{e}^x\right)+\frac{5}{2}{e}^{5x} \\ {y}^{\text{'}\text{'}}\left(x\right)=c_2{e}^x+c_3\left(x{e}^x+e^x+e^x\right)-12\left(x^2{e}^x+2x{e}^x+2\left(x{e}^x+e^x\right)\right)+\frac{25}{2}{e}^{5x} \\ {y}^{\text{'}\text{'}\text{'}}\left(x\right)=c_2{e}^x+c_3\left(x{e}^x+2e^x\right)-12\left(x^2{e}^x+4x{e}^x+2e^x\right)+\frac{25}{2}{e}^{5x} \end{gathered}$$

Applying initial condition we get:

$C_1=11, C_2=-11,C_3=9$

Now substitute the above values inthen

$$\begin{gathered} y\left(x\right)=c_1+c_2{e}^x+c_{3}x{e}^x+2x-12x^2{e}^x+\frac{1}{2}{e}^{5x} \\ y\left(x\right)=11-11e^x+9x{e}^x+2x-12x^2{e}^x+\frac{1}{2}{e}^{5x} \end{gathered}$$

Hence the solution is $\mathit{y}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathbf{11}\mathbf{-}\mathbf{11}{\mathit{e}}^{\mathbf{x}}\mathbf{+}\mathbf{9}\mathit{x}{\mathit{e}}^{\mathbf{x}}\mathbf{+}\mathbf{2}\mathit{x}\mathbf{-}\mathbf{12}{\mathit{x}}^{\mathbf{2}}{\mathit{e}}^{\mathbf{x}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{e}}^{\mathbf{5}\mathbf{x}}$