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Found in: Page 180

Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

Find a particular solution to the differential equation.$\frac{{\mathbf{d}}^{2}\mathbf{y}}{{\mathbf{dx}}^{2}}{\mathbf{-}}{\mathbf{5}}\frac{\mathrm{dy}}{\mathrm{dx}}{\mathbf{+}}{\mathbf{6}}{\mathbf{y}}{\mathbf{=}}{{\mathbf{xe}}}^{x}$

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

See the step by step solution

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

We have,

$\frac{{\mathbf{d}}^{2}\mathbf{y}}{{\mathbf{dx}}^{2}}\mathbf{-}\mathbf{5}\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{+}\mathbf{6}\mathbf{y}\mathbf{=}{\mathbf{xe}}^{x}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}......\left(\mathbf{1}\right)$

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,

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

The roots of the auxiliary equation are,

${\mathbf{m}}_{1}\mathbf{=}\mathbf{2}\mathbf{,}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}&\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathbf{m}}_{2}\mathbf{=}\mathbf{3}$

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),

${\mathrm{y}}_{\mathrm{p}}=\left(\mathrm{Ax}+\mathrm{B}\right){\mathrm{e}}^{\mathrm{x}}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}......\left(2\right)$

Now find the derivative of the above equation,

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

From the equation (1), Substitute the value of ${\mathrm{y}}_{\mathrm{p}}\text{'}\text{'},\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{y}}_{\mathrm{p}}\text{'}$ and ${\mathrm{y}}_{\mathrm{p}}$ in the equation (1),

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

Step 4: Final conclusion

Comparing all coefficients of the above equation;

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);

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

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