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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. ${\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}\left(\mathbf{x}\right){\mathbf{+}}{\mathbf{y}}\left(\mathbf{x}\right){\mathbf{=}}{{\mathbf{2}}}^{x}$

Thus, the particular solution is ${\mathrm{y}}_{\mathrm{p}}={2}^{\mathrm{x}}{\left({\left[\mathrm{ln}\left(2\right)\right]}^{2}+1\right)}^{-1}$

See the step by step solution

## Step 1: Use logarithms properties for simplification of the given differential equation.

The given differential equation is:

$\mathrm{y}\text{'}\text{'}\left(\mathrm{x}\right)+\mathrm{y}\left(\mathrm{x}\right)={2}^{\mathrm{x}}$

Simplify the above equation by using logarithms properties,

$\begin{array}{c}\mathrm{y}\text{'}\text{'}\left(\mathrm{x}\right)+\mathrm{y}\left(\mathrm{x}\right)={\mathrm{e}}^{\mathrm{ln}\left({2}^{\mathrm{x}}\right)}\\ \mathrm{y}\text{'}\text{'}\left(\mathrm{x}\right)+\mathrm{y}\left(\mathrm{x}\right)={\mathrm{e}}^{\mathrm{x}\left[\mathrm{ln}\left(2\right)\right]}\\ \mathrm{y}\text{'}\text{'}\left(\mathrm{x}\right)+\mathrm{y}\left(\mathrm{x}\right)={\mathrm{e}}^{\left[\mathrm{ln}\left(2\right)\right]\mathrm{x}}\end{array}$

## Step 2: Firstly, write the auxiliary equation of the above differential equation.

The auxiliary equation for the above equation:

${\mathrm{m}}^{2}+1=0$

## Step 3: Now find the roots of the auxiliary equation.

Solve the auxiliary equation,

$\begin{array}{c}{\mathrm{m}}^{2}+1=0\\ {\mathrm{m}}^{2}=-1\\ \mathrm{m}=\sqrt{-1}\\ \mathrm{m}=±\mathrm{i}\end{array}$

The roots of the auxiliary equation are:

${\mathbf{m}}_{1}\mathbf{=}\mathbf{i}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}&\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathbf{m}}_{2}\mathbf{=}\mathbf{-}\mathbf{i}$

The complementary solution of the given equation is:

${\mathbf{y}}_{c}\mathbf{=}{\mathbf{c}}_{1}\mathbf{cos}\left(\mathbf{x}\right)\mathbf{+}{\mathbf{c}}_{2}\mathbf{sin}\left(\mathbf{x}\right)$

## Step 4: Final conclusion, find a particular solution to the differential equation.

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

${\mathrm{y}}_{\mathrm{p}}={\mathrm{Ae}}^{\left[\mathrm{ln}\left(2\right)\right]\mathrm{x}}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\dots \left(2\right)$

Now find the derivative of the above equation,

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\text{'}=\mathrm{Aln}\left(2\right){\mathrm{e}}^{\left[\mathrm{ln}\left(2\right)\right]\mathrm{x}}\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}=\mathrm{A}{\left[\mathrm{ln}\left(2\right)\right]}^{2}{\mathrm{e}}^{\left[\mathrm{ln}\left(2\right)\right]\mathrm{x}}\end{array}$

From the equation (1),

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}+{\mathrm{y}}_{\mathrm{p}}={\mathrm{e}}^{\left[\mathrm{ln}\left(2\right)\right]\mathrm{x}}\\ \mathrm{A}{\left[\mathrm{ln}\left(2\right)\right]}^{2}{\mathrm{e}}^{\left[\mathrm{ln}\left(2\right)\right]\mathrm{x}}+{\mathrm{Ae}}^{\left[\mathrm{ln}\left(2\right)\right]\mathrm{x}}={\mathrm{e}}^{\left[\mathrm{ln}\left(2\right)\right]\mathrm{x}}\\ {\mathrm{Ae}}^{\left[\mathrm{ln}\left(2\right)\right]\mathrm{x}}\left[{\left[\mathrm{ln}\left(2\right)\right]}^{2}+1\right]={\mathrm{e}}^{\left[\mathrm{ln}\left(2\right)\right]\mathrm{x}}\\ \mathrm{A}\left[{\left[\mathrm{ln}\left(2\right)\right]}^{2}+1\right]=1\\ \mathrm{A}=\frac{}{{\left[\mathrm{ln}\left(2\right)\right]}^{2}+1}\end{array}$

Substitute the value of A in the equation (2), we get:

${\mathrm{y}}_{\mathrm{p}}={2}^{\mathrm{x}}{\left({\left[\mathrm{ln}\left(2\right)\right]}^{2}+1\right)}^{-1}$

Therefore, the particular solution of equation (1),

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