Q7E - 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. $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}\mathbf{2}\mathbf{y}\mathbf{+}\mathbf{2}{\mathbf{tan}}^3\mathbf{x}\mathbf{,}{\mathbf{y}}_p\left(\mathbf{x}\right)\mathbf{=}\mathbf{tanx}$

Step-by-Step Solution

Answer

The general solution of the given differential equation is $\mathrm{y}={\mathrm{c}}_1{\mathrm{e}}^{\sqrt{2}\mathrm{x}}+{\mathrm{c}}_2{\mathrm{e}}^{-\sqrt{2}\mathrm{x}}+\mathrm{tanx}$.

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

The differential equation is,

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

 

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,

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

 

The roots of the auxiliary equation are, 

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

 

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,

$\begin{array}{c}\mathrm{y}={\mathrm{y}}_{\mathrm{c}}\left(\mathrm{x}\right)+{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)\\ \mathrm{y}={\mathrm{c}}_1{\mathrm{e}}^{\sqrt{2}\mathrm{x}}+{\mathrm{c}}_2{\mathrm{e}}^{-\sqrt{2}\mathrm{x}}+\mathrm{tanx}\end{array}$