Q9E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

find a general solution to the given equation.

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

Step-by-Step Solution

Answer

$y\left(x\right)=\frac{1}{6}{x}^3{e}^x+c_1{e}^x+c_{2}x{e}^x+c_3{x}^2{e}^x$

Step 1: Find the corresponding auxiliaryequation

Theauxiliary equationof corresponding homogeneous equation 

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

The solutions of the auxiliary equation are

$r=1,r=1,r=1$

Therefore a general solution to the homogeneous equation is

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

Step 2: Find particular solution

Let the particular solution be

$y_p\left(x\right)=a{x}^3{e}^x$

 Then

$$\begin{gathered} y_p^{\text{'}}\left(x\right)=3a{x}^2{e}^x+a{x}^3{e}^x \\ {y}_p^{\text{'}\text{'}}\left(x\right)=6ax{e}^x+6a{x}^2{e}^x+a{x}^3{e}^x \\ {y}_p^{\text{'}\text{'}\text{'}}\left(x\right)=6a{e}^x+18ax{e}^x+9a{x}^2{e}^x+a{x}^3{e}^x \end{gathered}$$

Then

$$\begin{gathered} y_p^{\text{'}\text{'}\text{'}}\left(x\right)-3y_p^{\text{'}\text{'}}\left(x\right)+3y_p^{\text{'}\text{'}}\left(x\right)-y_p^{\text{'}}\left(x\right) \\ =6a{e}^x+18ax{e}^x+9a{x}^2{e}^x+a{x}^3{e}^x-18ax{e}^x-18a{x}^2{e}^x-3a{x}^3{e}^x+9a{x}^2{e}^x+3a{x}^3{e}^x-a{x}^3{e}^x\backslash \mathrm{hfill} \\ =6a{e}^x \end{gathered}$$

If $6a{e}^x=e^x$

Then $a=\frac{1}{6}$

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

 

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

Then $y\left(x\right)=\frac{1}{6}{x}^3{e}^x+c_1{e}^x+c_{2}x{e}^x+c_3{x}^2{e}^x$

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