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Expert-verified 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.${\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{2}}{\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{-}}{\mathbf{y}}{\mathbf{=}}{\mathbf{10}}$

Thus, the particular solution is ${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)=-10.$

See the step by step solution

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

Given the differential equation,

$\mathrm{y}\text{'}\text{'}+2\mathrm{y}\text{'}-\mathrm{y}=10\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\dots \left(1\right)$

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

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

The auxiliary equation for the above equation,

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

## Step 2: Now find the roots of the auxiliary equation.

Solve the auxiliary equation,

$\begin{array}{c}{\mathrm{m}}^{2}+2\mathrm{m}-1=0\\ \mathrm{m}=\frac{-2±\sqrt{{2}^{2}-4\left(1\right)\left(-1\right)}}{2\left(1\right)}\\ \mathrm{m}=\frac{-2±\sqrt{8}}{2}\\ \mathrm{m}=-1±\sqrt{2}\end{array}$

The roots of the auxiliary equation are:

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

The complementary solution of the given equation is:

${\mathrm{y}}_{\mathrm{c}}\left(\mathrm{x}\right)={\mathrm{e}}^{-\mathrm{t}}\left[{\mathrm{c}}_{1}\mathrm{cosh}\left(\sqrt{2}\mathrm{t}\right)+{\mathrm{c}}_{2}\mathrm{sinh}\left(\sqrt{2}\mathrm{t}\right)\right]$

## Step 3: Final conclusion, find a particular solution to the differential equation.

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

${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)=\mathrm{A}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\dots \left(2\right)$

Now find the derivative of the above equation,

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{x}\right)=0\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{x}\right)=0\end{array}$

From the equation (1),

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}+2{\mathrm{y}}_{\mathrm{p}}\text{'}-{\mathrm{y}}_{\mathrm{p}}=10\\ \left(0\right)+2\left(0\right)-\mathrm{A}=10\\ \mathrm{A}=-10\end{array}$

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

${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)=-10.$

Therefore, the particular solution of the differential equation is:

${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)=-10$ ### Want to see more solutions like these? 