Q8E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

A nonhomogeneous equation and a particular solution are given. Find a general solution for the equation.

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}\mathbf{2}\mathit{y}\mathbf{\text{'}}\mathbf{-}\mathit{y}\mathbf{+}\mathbf{2}{\mathit{e}}^{\mathit{x}}\mathbf{,} {\mathit{y}}_{\mathit{p}}\left(\mathit{x}\right)\mathbf{=}{\mathit{x}}^{\mathbf{2}}{\mathit{e}}^{\mathit{x}}$

Step-by-Step Solution

Answer

The general solution of the given differential equation is $y=c_1{e}^x+c_{2}x{e}^x+x^2{e}^x$

Step 1: Write the auxiliary equation of the given differential equation.

The differential equation is,

 $$\begin{gathered} y\text{'}\text{'}=2y\text{'}-y+2e^x \\ y\text{'}\text{'}-2y\text{'}+y=2e^x \dots \left(1\right) \end{gathered}$$

 

Write the homogeneous differential equation of the equation (1),

$y\text{'}\text{'}-2y\text{'}+y=0$ 

 

The auxiliary equation for the above equation,

$m^2-2m+1=0$

Step 2: Now find the complementary solution of the given equation.

Solve the auxiliary equation,

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

 

The roots of the auxiliary equation are, 

$m_1=1, m_2=1$

 

The complementary solution of the given equation is,

$y_c=c_1{e}^x+c_{2}x{e}^x$

Step 3: Use the given particular solution to find a general solution for the equation.

The given particular solution,

 $y_p\left(x\right)=x^2{e}^x$

 

Therefore, the general solution is,

$$\begin{gathered} y=y_c\left(x\right)+y_p\left(x\right) \\ y=c_1{e}^x+c_{2}x{e}^x+x^2{e}^x \end{gathered}$$