Q7E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

find a general solution to the given equation.

$y\text{'}\text{'}\text{'}+3y\text{'}\text{'}-4y=e^{-2x}$

Step-by-Step Solution

Answer

$y\left(x\right)=-\frac{1}{6}{x}^2{e}^{-2x}+c_1{e}^{-2x}+c_{2}x{e}^{-2x}+c_3{e}^x$

Step 1: Find the corresponding auxiliaryequation

Theauxiliary equationof corresponding homogeneous equation 

$r^3+3r^2-4=(r-1)(r+2{)}^2=0$

The solutions of the auxiliary equation are

$r=-2,r=-2,r=1$

Therefore a general solution to the homogeneous equation is

$y_h\left(x\right)=c_1{e}^{-2x}+c_{2}x{e}^{-2x}+c_3{e}^x$

Step 2: Find particular solution

Let the particular solution be

$y_p\left(x\right)=a{x}^2{e}^{-2x}$

 Then

$$\begin{gathered} y_p^{\text{'}}\left(x\right)=2ax{e}^{-2x}-2a{x}^2{e}^{-2x} \\ {y}_p^{\text{'}\text{'}}\left(x\right)=2a{e}^{-2x}-8ax{e}^{-2x}+4a{x}^2{e}^{-2x} \\ {y}_p^{\text{'}\text{'}\text{'}}\left(x\right)=-12a{e}^{-2x}+24ax{e}^{-2x}-8a{x}^2{e}^{-2x} \end{gathered}$$

Then

$$\begin{gathered} y_p^{\text{'}\text{'}\text{'}}\left(x\right)+3y_p^{\text{'}\text{'}}\left(x\right)-4y_p\left(x\right) \\ =-12a{e}^{-2x}+24ax{e}^{-2x}-8a{x}^2{e}^{-2x}+6a{e}^{-2x}-24ax{e}^{-2x}+12a{x}^2{e}^{-2x}-4a{x}^2{e}^{-2x} \\ =-6a{e}^{-2x}. \end{gathered}$$

If $-6a{e}^{-2x}=e^{-2x } then 6a=1$

Then $a=-\frac{1}{6}$

Hence

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

Step 3: y ( x ) = y h + y p

Then $y\left(x\right)=-\frac{1}{6}{x}^2{e}^{-2x}+c_1{e}^{-2x}+c_{2}x{e}^{-2x}+c_3{e}^x$

Is the general solution of $y\text{'}\text{'}\text{'}+3y\text{'}\text{'}-4y=e^{-2x}$