Question

In Problemssolve the given initial-value problem.

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathit{y}\mathbf{=}\mathbf{(}\mathbf{3}\mathbf{+}\mathit{x}\mathbf{)}{\mathit{e}}^{\mathbf{-}\mathbf{2}\mathbf{x}}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{,}\mathit{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{5}\mathbf{ }$

Short answer

$y=e^{-2x}\left(2+9x+\frac{1}{6}{x}^3+\frac{3}{2}{x}^2\right)$

Step-by-step solution

Complementary function

Consider the differential equation$y\text{'}\text{'}+4y\text{'}+4y=(3+x)e^{-2x},y\left(0\right)=2,y\text{'}\left(0\right)=5 ······\left(1\right)$

The auxiliary equation corresponding to homogenous equation is:

$$\begin{gathered} m^2+4m+4=0 \\ (m+2)(m+2)=0 \\ m=-2,-2 \end{gathered}$$

So, the roots are real and equal.

Thus, the complementary function is$y_c=C_1{e}^{-2x}+C_{2}x{e}^{-2x}$

Form particular solution

For particular solution assume$y_p=\left(A{x}^3+B{x}^2\right)e^{-2x}$

Differentiating we have:

$$\begin{gathered} y_p\text{'}=-2\left(A{x}^3+B{x}^2\right)e^{-2x}+\left(3A{x}^2+2Bx\right)e^{-2x} \\ {y}_p\text{'}\text{'}=4\left(A{x}^3+B{x}^2\right)e^{-2x}-4\left(3A{x}^2+2Bx\right)e^{-2x}+6Ax{e}^{-2x}+2B{e}^{-2x} \end{gathered}$$

Substituting values in equation (1) we have:

$$\begin{gathered} 4\left(A{x}^3+B{x}^2\right)e^{-2x}-4\left(3A{x}^2+2Bx\right)e^{-2x}+6Ax{e}^{-2x}+2B{e}^{-2x} \\ +4\left[-2\left(A{x}^3+B{x}^2\right)e^{-2x}+\left(3A{x}^2+2Bx\right)e^{-2x}\right]+4\left[\left(A{x}^3+B{x}^2\right)e^{-2x}\right]=(3+x)e^{-2x} \\ 4A{x}^3{e}^{-2x}+4B{x}^2{e}^{-2x}-12A{x}^2{e}^{-2x}-8Bx{e}^{-2x}+6Ax{e}^{-2x}+2B{e}^{-2x} \\ -8A{x}^3{e}^{-2x}-8B{x}^2{e}^{-2x}+12A{x}^2{e}^{-2x}+8Bx{e}^{-2x}+4A{x}^3{e}^{-2x}+4B{x}^2{e}^{-2x}=3e^{-2x}+x{e}^{-2x} \\ (4A-8A+4A)x^3{e}^{-2x}+(4B-12A-8B+12A+4B)x^2{e}^{-2x}+(-8B+6A+8B)x{e}^{-2x}+2B{e}^{-2x}=3e^{-2x}+x{e}^{-2x} \\ 6Ax{e}^{-2x}+2B{e}^{-2x}=x{e}^{-2x}+3e^{-2x} \end{gathered}$$

General solution

On equating the coefficients of like terms, obtain

$$\begin{gathered} 6A=1\Rightarrow A=\frac{1}{6}x{e}^{-2x}\text{coefficients} \\ 2B=3\Rightarrow B=\frac{3}{2}{e}^{-2x}\text{coefficients} \end{gathered}$$

So, particular solution is.$y_p=\left(A{x}^3+B{x}^2\right)e^{-2x}$

$y_p=\left(\frac{1}{6}{x}^3+\frac{3}{2}{x}^2\right)e^{-2x}$

Thus, the solution is:

$$\begin{gathered} y=y_c+y_p \\ y\left(x\right)=C_1{e}^{-2x}+C_{2}x{e}^{-2x}+\left(\frac{1}{6}{x}^3+\frac{3}{2}{x}^2\right)e^{-2x}······\left(2\right) \end{gathered}$$

Apply initial condition

Apply $y\text{'}\left(0\right)=5$

$$\begin{gathered} y\text{'}\left(0\right)=-2C_1{e}^{-2\left(0\right)}-2C_2\left(0\right)e^{-2\left(0\right)}+C_2{e}^{-2\left(0\right)}-2\left(\frac{1}{6}{\left(0\right)}^3+\frac{3}{2}{\left(0\right)}^2\right)e^{-2\left(0\right)}+\left(\frac{1}{2}{\left(0\right)}^2+3\left(0\right)\right)e^{-2\left(0\right)} \\ 5=-2C_1+0+C_2-0+0 \\ 5=-2\left(2\right)+C_2\text{Since}{C}_1=2 \\ {C}_2=9 \end{gathered}$$

Substitute localid="1663919948179" $C_1=2and{C}_2=9$in the solution (2).

$y\left(x\right)=2e^{-2x}+9x{e}^{-2x}+\left(\frac{1}{6}{x}^3+\frac{3}{2}{x}^2\right)e^{-2x}$

Hence, the general solution of the given initial value problem is,

$\mathit{y}\mathbf{=}{\mathit{e}}^{\mathbf{-}\mathbf{2}\mathbf{x}}\left(2+9x+\frac{1}{6}{x}^3+\frac{3}{2}{x}^2\right)$