Q3E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

A nonhomogeneous equation and a particular solution are given. Find a general solution for the equation.$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{=}\mathbf{t}\mathbf{,}{\mathbf{y}}_p\left(\mathbf{t}\right)\mathbf{=}\mathbf{-}\mathbf{t}$

Step-by-Step Solution

Answer

The general solution of the differential equation is $\mathrm{y}={\mathrm{c}}_1{\mathrm{e}}^{\mathrm{t}}+{\mathrm{c}}_2{\mathrm{e}}^{-\mathrm{t}}-\mathrm{t}.$

Step 1: Write the auxiliary equation of the given differential equation.

The differential equation is,

$\mathrm{y}\text{'}\text{'}-\mathrm{y}=\mathrm{t}. .....\left(1\right)$

 

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

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

 

The auxiliary equation for the above equation,

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

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

Solve the auxiliary equation,

$\begin{array}{c}{\mathrm{m}}^2-1=0\\ \mathrm{m}=\pm 1\end{array}$

 

The roots of the auxiliary equation are, 

${\mathrm{m}}_1=1,{\mathrm{m}}_2=-1$

The complementary solution of the given equation is,

 ${\mathrm{y}}_{\mathrm{c}}={\mathrm{c}}_1{\mathrm{e}}^{\mathrm{t}}+{\mathrm{c}}_2{\mathrm{e}}^{-\mathrm{t}}$

 

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

The given particular solution,

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

 

Therefore, the general solution is,

$\begin{array}{c}\mathrm{y}={\mathrm{y}}_{\mathrm{c}}\left(\mathrm{t}\right)+{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)\\ \mathrm{y}={\mathrm{c}}_1{\mathrm{e}}^{\mathrm{t}}+{\mathrm{c}}_2{\mathrm{e}}^{-\mathrm{t}}-\mathrm{t}\end{array}$