Q6E - 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.$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{6}\mathbf{y}\mathbf{=}\mathbf{6}{\mathbf{x}}^2\mathbf{+}\mathbf{10}\mathbf{x}\mathbf{+}\mathbf{2}\mathbf{+}\mathbf{12}{\mathbf{e}}^x\mathbf{,}{\mathbf{y}}_p\left(\mathbf{x}\right)\mathbf{=}{\mathbf{e}}^x\mathbf{+}{\mathbf{x}}^2$

Step-by-Step Solution

Answer

The general solution of the given differential equation is  $\mathrm{y}={\mathrm{c}}_1{\mathrm{e}}^{-2\mathrm{x}}+{\mathrm{c}}_2{\mathrm{e}}^{-3\mathrm{x}}+{\mathrm{e}}^{\mathrm{x}}+{\mathrm{x}}^2$.

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

The differential equation is,

 

$\mathrm{y}\text{'}\text{'}+5\mathrm{y}\text{'}+6\mathrm{y}=6{\mathrm{x}}^2+10\mathrm{x}+2+12{\mathrm{e}}^{\mathrm{x}}\dots \left(1\right)$

 

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

 

$\mathrm{y}\text{'}\text{'}+5\mathrm{y}\text{'}+6\mathrm{y}=0$

 

The auxiliary equation for the above equation,

 ${\mathrm{m}}^2+5\mathrm{m}+6=0$

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

Solve the auxiliary equation,

 $\begin{array}{c}{\mathrm{m}}^2+5\mathrm{m}+6=0\\ {\mathrm{m}}^2+3\mathrm{m}+2\mathrm{m}+6=0\\ \mathrm{m}(\mathrm{m}+3)+2(\mathrm{m}+3)=0\\ (\mathrm{m}+2)(\mathrm{m}+3)=0\end{array}$

 

The roots of the auxiliary equation are, 

${\mathrm{m}}_1=-2,\&{\mathrm{m}}_2=-3$

 

The complementary solution of the given equation is,

 ${\mathrm{y}}_{\mathrm{c}}={\mathrm{c}}_1{\mathrm{e}}^{-2\mathrm{x}}+{\mathrm{c}}_2{\mathrm{e}}^{-3\mathrm{x}}$

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

The given particular solution,

${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)={\mathrm{e}}^{\mathrm{x}}+{\mathrm{x}}^2$

Therefore, the general solution is,

$\begin{array}{c}\mathrm{y}={\mathrm{y}}_{\mathrm{c}}\left(\mathrm{x}\right)+{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)\\ \mathrm{y}={\mathrm{c}}_1{\mathrm{e}}^{-2\mathrm{x}}+{\mathrm{c}}_2{\mathrm{e}}^{-3\mathrm{x}}+{\mathrm{e}}^{\mathrm{x}}+{\mathrm{x}}^2\end{array}$