Q18E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find a particular solution to the differential equation.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{8}{\mathbf{e}}^t$

Step-by-Step Solution

Answer

The particular solution to the given differential equation is ${\mathbf{y}}_p\left(\mathbf{x}\right)\mathbf{=}\mathbf{4}{\mathbf{t}}^2{\mathbf{e}}^t$.

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

The given differential equation,

 

$\mathrm{y}\text{'}\text{'}-2\mathrm{y}\text{'}+\mathrm{y}=8{\mathrm{e}}^{\mathrm{t}}\dots \left(1\right)$

 

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,

 

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

 

The roots of the auxiliary equation are, 

 

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

 

The complementary solution of the given equation is,

 

${\mathrm{y}}_{\mathrm{c}}\left(\mathrm{x}\right)={\mathrm{c}}_1{\mathrm{e}}^{\mathrm{t}}+{\mathrm{c}}_2{\mathrm{te}}^{\mathrm{t}}$

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),

${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)={\mathrm{At}}^2{\mathrm{e}}^{\mathrm{t}}\dots \left(2\right)$ 

 

Now find the derivative of the above equation,

 $\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{x}\right)=({\mathrm{At}}^2+2\mathrm{At}){\mathrm{e}}^{\mathrm{t}}\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{x}\right)=({\mathrm{At}}^2+4\mathrm{At}+2\mathrm{A}){\mathrm{e}}^{\mathrm{t}}\end{array}$

 

From the equation (1),

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}-2{\mathrm{y}}_{\mathrm{p}}\text{'}+{\mathrm{y}}_{\mathrm{p}}=8{\mathrm{e}}^{\mathrm{t}}\\ ({\mathrm{At}}^2+4\mathrm{At}+2\mathrm{A}){\mathrm{e}}^{\mathrm{t}}-2\left(({\mathrm{At}}^2+2\mathrm{At}){\mathrm{e}}^{\mathrm{t}}\right)+{\mathrm{At}}^2{\mathrm{e}}^{\mathrm{t}}=8{\mathrm{e}}^{\mathrm{t}}\\ \left(2\mathrm{A}\right){\mathrm{e}}^{\mathrm{t}}=8{\mathrm{e}}^{\mathrm{t}}\end{array}$

Step 4: Final conclusion:

Comparing all coefficients of the above equation;

 $\begin{array}{c}2\mathrm{A}=8\\ \mathrm{A}=4\end{array}$

 

Therefore, the particular solution of equation (1),

 ${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)=4{\mathrm{t}}^2{\mathrm{e}}^{\mathrm{t}}$