Q10E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find a particular solution to the differential equation.

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

Step-by-Step Solution

Answer

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

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

Given the differential equation,

 

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

 

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,

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

 

The roots of the auxiliary equation are:

 

${\mathrm{m}}_1=-1+\sqrt{2}\&{\mathrm{m}}_2=-1-\sqrt{2}$

 

The complementary solution of the given equation is:

 

${\mathrm{y}}_{\mathrm{c}}\left(\mathrm{x}\right)={\mathrm{e}}^{-\mathrm{t}}[{\mathrm{c}}_1\cosh \left(\sqrt{2}\mathrm{t}\right)+{\mathrm{c}}_2\sinh \left(\sqrt{2}\mathrm{t}\right)]$

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),

 

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

 

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:

 

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