Question

In Problems 1–26 solve the given differential equation by undetermined coefficients.

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{4}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathit{y}\mathbf{=}\mathbf{5}\mathbf{-}{\mathit{e}}^{\mathbf{x}}\mathbf{+}{\mathit{e}}^{\mathbf{2}\mathbf{x}}$

Short answer

The solution is$y=C_1{e}^x+C_2{e}^{2x}+C_3{e}^{-2x}+\frac{5}{4}+\frac{1}{3}x{e}^x+\frac{1}{4}x{e}^{2x}$

Step-by-step solution

Form complementary function

Consider the following differential equation:$y\text{'}\text{'}\text{'}-y\text{'}\text{'}-4y\text{'}+4y=5-e^x+e^{2x}$

The auxiliary equation of the homogeneous differential equation :

$$\begin{gathered} y\text{'}\text{'}\text{'}-y\text{'}\text{'}-4y\text{'}+4y=0 \\ {m}^3-m^2-4m+4=0 \end{gathered}$$

Clearly is a root of the equation.

So, the above cubic equation becomes,

$$\begin{gathered} (m-1)\left(m^2-4\right)=0 \\ (m-1)(m-2)(m+2)=0 \\ m=1,2,-2 \end{gathered}$$

Therefore, the complementary function is$y_c\left(x\right)=C_1{e}^x+C_2{e}^{2x}+C_3{e}^{-2x}$

Find particular solution

Let the particular integral be$y_p\left(x\right)=A+Bx{e}^x+Cx{e}^{2x}.$

Then,

$$\begin{gathered} y_p\text{'}\left(x\right)=B{e}^x+Bx{e}^x+C{e}^{2x}+2Cx{e}^{2x} \\ {y}_p\text{'}\text{'}\left(x\right)=2B{e}^x+Bx{e}^x+4C{e}^{2x}+4Cx{e}^{2x} \\ {y}_p\text{'}\text{'}\text{'}\left(x\right)=3B{e}^x+Bx{e}^x+12C{e}^{2x}+8Cx{e}^{2x} \end{gathered}$$

Substitute these values in the given differential equation to obtain that,

$$\begin{gathered} y_p^{\text{'}\text{'}\text{'}}-y_p^{\text{'}\text{'}}-4y_p+4y_p=5-e^x+e^{2x}\left(3B{e}^x+Bx{e}^x+12C{e}^{2x}+8Cx{e}^{2x}\right)-\left(2B{e}^x+Bx{e}^x+4C{e}^{2x}+4Cx{e}^{2x}\right)-4\left(B{e}^x+Bx{e}^x+C{e}^{2x}+2Cx{e}^{2x}\right)+4\left(A+Bx{e}^x+Cx{e}^{2x}\right)=5-e^x+e^{2x} \\ 4A+e^x(3B-2B-4B)+e^{2x}(12C-4C-4C)+x{e}^x(B-B-4B+4B)+x{e}^{2x}(8C-4C-8C+4C)=5-e^x+e^{2x} \end{gathered}$$

General solution

Compare the like terms we have:

$$\begin{gathered} 4A=5,-3B=-1,4C=1 \\ \Rightarrow A=\frac{5}{4},B=\frac{1}{3},C=\frac{1}{4} \end{gathered}$$

Therefore, the particular integral is,$y_p\left(x\right)=\frac{5}{4}+\frac{1}{3}x{e}^x+\frac{1}{4}x{e}^{2x}.$

Thus, the general solution is,

$$\begin{gathered} \mathit{y}\mathbf{=}{\mathit{y}}_{\mathbf{c}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{+}{\mathit{y}}_{\mathbf{p}}\mathbf{\left(}\mathit{x}\mathbf{\right)} \\ \mathit{y}\mathbf{=}{\mathit{C}}_{\mathbf{1}}{\mathit{e}}^{\mathbf{x}}\mathbf{+}{\mathit{C}}_{\mathbf{2}}{\mathit{e}}^{\mathbf{2}\mathbf{x}}\mathbf{+}{\mathit{C}}_{\mathbf{3}}{\mathit{e}}^{\mathbf{-}\mathbf{2}\mathbf{x}}\mathbf{+}\frac{\mathbf{5}}{\mathbf{4}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{3}}\mathit{x}{\mathit{e}}^{\mathbf{x}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{4}}\mathit{x}{\mathit{e}}^{\mathbf{2}\mathbf{x}} \end{gathered}$$