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### 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

# A nonhomogeneous equation and a particular solution are given. Find a general solution for the equation. ${\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{=}}{\mathbf{2}}{\mathbf{y}}{\mathbf{+}}{\mathbf{2}}{{\mathbf{tan}}}^{3}{\mathbf{x}}{\mathbf{,}}{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}}{{\mathbf{y}}}_{p}\left(\mathbf{x}\right){\mathbf{=}}{\mathbf{tanx}}$

The general solution of the given differential equation is $\mathrm{y}={\mathrm{c}}_{1}{\mathrm{e}}^{\sqrt{2}\mathrm{x}}+{\mathrm{c}}_{2}{\mathrm{e}}^{-\sqrt{2}\mathrm{x}}+\mathrm{tanx}$.

See the step by step solution

## Step 1: Firstly, write the auxiliary equation of the given differential equation

The differential equation is,

$\begin{array}{c}\mathrm{y}\text{'}\text{'}=2\mathrm{y}+2{\mathrm{tan}}^{3}\mathrm{x}\\ \mathrm{y}\text{'}\text{'}-2\mathrm{y}=2{\mathrm{tan}}^{3}\mathrm{x}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\dots \left(1\right)\end{array}$

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

$\mathrm{y}\text{'}\text{'}-2\mathrm{y}=0$

The auxiliary equation for the above equation,

${\mathrm{m}}^{2}-2=0$

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

Solve the auxiliary equation,

$\begin{array}{c}{\mathrm{m}}^{2}-2=0\\ \mathrm{m}=±\sqrt{2}\end{array}$

The roots of the auxiliary equation are,

${\mathrm{m}}_{1}=\sqrt{2},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}&\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{m}}_{2}=-\sqrt{2}$

The complementary solution of the given equation is,

${\mathrm{y}}_{\mathrm{c}}={\mathrm{c}}_{1}{\mathrm{e}}^{\sqrt{2}\mathrm{x}}+{\mathrm{c}}_{2}{\mathrm{e}}^{-\sqrt{2}\mathrm{x}}$

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

The given particular solution,

${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)=\mathrm{tanx}$

Therefore, the general solution is,

$\begin{array}{c}\mathrm{y}={\mathrm{y}}_{\mathrm{c}}\left(\mathrm{x}\right)+{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)\\ \mathrm{y}={\mathrm{c}}_{1}{\mathrm{e}}^{\sqrt{2}\mathrm{x}}+{\mathrm{c}}_{2}{\mathrm{e}}^{-\sqrt{2}\mathrm{x}}+\mathrm{tanx}\end{array}$

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