Q15E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find a particular solution to the differential equation.

$\frac{{\mathbf{d}}^2\mathbf{y}}{{\mathbf{dx}}^2}\mathbf{-}\mathbf{5}\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{+}\mathbf{6}\mathbf{y}\mathbf{=}{\mathbf{xe}}^x$

Step-by-Step Solution

Answer

The particular solution of the differential equation is ${\mathrm{y}}_{\mathrm{p}}=(\frac{1}{2}\mathrm{x}+\frac{3}{4}){\mathrm{e}}^{\mathrm{x}}.$

Step 1: Firstly, write the auxiliary equation of the above differential equation:

We have,

 

$\frac{{\mathbf{d}}^2\mathbf{y}}{{\mathbf{dx}}^2}\mathbf{-}\mathbf{5}\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{+}\mathbf{6}\mathbf{y}\mathbf{=}{\mathbf{xe}}^x\mathrm{......}\left(\mathbf{1}\right)$

 

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

 $\frac{{\mathbf{d}}^2\mathbf{y}}{{\mathbf{dx}}^2}\mathbf{-}\mathbf{5}\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{+}\mathbf{6}\mathbf{y}\mathbf{=}\mathbf{0}$

The auxiliary equation for the above equation,

${\mathbf{m}}^2\mathbf{-}\mathbf{5}\mathbf{m}\mathbf{+}\mathbf{6}\mathbf{=}\mathbf{0}$

 

Step 2: Now find the roots of the auxiliary equation

Solve the auxiliary equation,

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

The roots of the auxiliary equation are, 

 

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

 

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 method of undetermined coefficients to find a particular solution to the differential equation

According to the method of undetermined coefficients, assume, the particular solution of equation (1),

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

 

Now find the derivative of the above equation,

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\text{'}=\left(\mathrm{A}\right){\mathrm{e}}^{\mathrm{x}}+(\mathrm{Ax}+\mathrm{B}){\mathrm{e}}^{\mathrm{x}}\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}=\left(\mathrm{A}\right){\mathrm{e}}^{\mathrm{x}}+\left(\mathrm{A}\right){\mathrm{e}}^{\mathrm{x}}+(\mathrm{Ax}+\mathrm{B}){\mathrm{e}}^{\mathrm{x}}\\ =2\left(\mathrm{A}\right){\mathrm{e}}^{\mathrm{x}}+(\mathrm{Ax}+\mathrm{B}){\mathrm{e}}^{\mathrm{x}}\end{array}$

From the equation (1), Substitute the value of ${\mathrm{y}}_{\mathrm{p}}\text{'}\text{'},{\mathrm{y}}_{\mathrm{p}}\text{'}$ and ${\mathrm{y}}_{\mathrm{p}}$ in the equation (1),

$\begin{array}{c}\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}-5\frac{}{}+6\mathrm{y}={\mathrm{xe}}^{\mathrm{x}}\\ 2\left(\mathrm{A}\right){\mathrm{e}}^{\mathrm{x}}+(\mathrm{Ax}+\mathrm{B}){\mathrm{e}}^{\mathrm{x}}-5(\left(\mathrm{A}\right){\mathrm{e}}^{\mathrm{x}}+(\mathrm{Ax}+\mathrm{B}){\mathrm{e}}^{\mathrm{x}})+6\left((\mathrm{Ax}+\mathrm{B}){\mathrm{e}}^{\mathrm{x}}\right)={\mathrm{xe}}^{\mathrm{x}}\\ -3\left(\mathrm{A}\right){\mathrm{e}}^{\mathrm{x}}+2(\mathrm{Ax}+\mathrm{B}){\mathrm{e}}^{\mathrm{x}}={\mathrm{xe}}^{\mathrm{x}}\\ 2{\mathrm{Axe}}^{\mathrm{x}}+(-3\mathrm{A}+2\mathrm{B}){\mathrm{e}}^{\mathrm{x}}={\mathrm{xe}}^{\mathrm{x}}\end{array}$

 

Step 4: Final conclusion

Comparing all coefficients of the above equation;

 $\begin{array}{c}2\mathrm{A}=1\\ \mathrm{A}=\frac{1}{2}\\ -3\mathrm{A}+2\mathrm{B}=0......\left(3\right)\end{array}$

 

Substitute the value of Ain the equation (3),

 

$\begin{array}{c}-3\left(\frac{1}{2}\right)+2\mathrm{B}=0\\ \mathrm{B}=\frac{3}{4}\end{array}$

 

Substitute the value of A and B in the equation (2),

 ${\mathrm{y}}_{\mathrm{p}}=\left(\frac{1}{2}\mathrm{x}+\frac{3}{4}\right){\mathrm{e}}^{\mathrm{x}}$

 

Therefore, the particular solution of equation (1);

${\mathrm{y}}_{\mathrm{p}}=\left(\frac{1}{2}\mathrm{x}+\frac{3}{4}\right){\mathrm{e}}^{\mathrm{x}}$