Q19E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find a particular solution to the differential equation.$\mathbf{4}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{11}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{3}\mathbf{y}\mathbf{=}\mathbf{-}\mathbf{2}{\mathbf{te}}^{-3t}$

Step-by-Step Solution

Answer

The particular solution is ${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)=\left(\frac{1}{13}{\mathrm{t}}^2+\frac{8}{169}\mathrm{t}\right){\mathrm{e}}^{-3\mathrm{t}}$

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

The differential equation is,

 

$4\mathrm{y}\text{'}\text{'}+11\mathrm{y}\text{'}-3\mathrm{y}=-2{\mathrm{te}}^{-3\mathrm{t}}\dots \left(1\right)$

 

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

 

$4\mathrm{y}\text{'}\text{'}+11\mathrm{y}\text{'}-3\mathrm{y}=0$

 

The auxiliary equation for the above equation,

 

$4{\mathrm{m}}^2+11\mathrm{m}-3=0$

Step 2: Now find the roots of the auxiliary equation

Solve the auxiliary equation,

$\begin{array}{c}4{\mathrm{m}}^2+11\mathrm{m}-3=0\\ 4{\mathrm{m}}^2+12\mathrm{m}-\mathrm{m}-3=0\\ 4\mathrm{m}(\mathrm{m}+3)-1(\mathrm{m}+3)=0\\ (\mathrm{m}+3)(4\mathrm{m}-1)=0\end{array}$ 

 

The roots of the auxiliary equation are, 

 

${\mathbf{m}}_1\mathbf{=}\mathbf{-}\mathbf{3}\mathbf{,}\&{\mathbf{m}}_2\mathbf{=}\frac{1}{4}$

 

The complementary solution of the given equation is,

 ${\mathbf{y}}_c\left(\mathbf{x}\right)\mathbf{=}{\mathbf{c}}_1{\mathbf{e}}^{-3t}\mathbf{+}{\mathbf{c}}_2{\mathbf{e}}^{\frac{t}{4}}$

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),

${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)=({\mathrm{At}}^2+\mathrm{Bt}){\mathrm{e}}^{-3\mathrm{t}}\dots \left(2\right)$

Now find the derivative of the above equation,

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{x}\right)=({\mathrm{At}}^2+\mathrm{Bt}){\mathrm{e}}^{-3\mathrm{t}}(-3)+(2\mathrm{At}+\mathrm{B}){\mathrm{e}}^{-3\mathrm{t}}\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{x}\right)=(-3{\mathrm{At}}^2-3\mathrm{Bt}+2\mathrm{At}+\mathrm{B}){\mathrm{e}}^{-3\mathrm{t}}\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{x}\right)=(-6\mathrm{At}-3\mathrm{B}+2\mathrm{A}){\mathrm{e}}^{-3\mathrm{t}}+(-3{\mathrm{At}}^2-3\mathrm{Bt}+2\mathrm{At}+\mathrm{B}){\mathrm{e}}^{-3\mathrm{t}}(-3)\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{x}\right)=(9{\mathrm{At}}^2-12\mathrm{At}+9\mathrm{Bt}+2\mathrm{A}-6\mathrm{B}){\mathrm{e}}^{-3\mathrm{t}}\end{array}$ 

 

From the equation (1), Substitute the value of  ${\mathrm{y}}_{\mathrm{p}}\text{'}\text{'},{\mathrm{y}}_{\mathrm{p}}\text{'}$and ${\mathbf{y}}_p$ in the equation (1),

$\begin{array}{c}4{\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}+11{\mathrm{y}}_{\mathrm{p}}\text{'}-3{\mathrm{y}}_{\mathrm{p}}=-2{\mathrm{te}}^{-3\mathrm{t}}\\ 4(9{\mathrm{At}}^2-12\mathrm{At}+9\mathrm{Bt}+2\mathrm{A}-6\mathrm{B}){\mathrm{e}}^{-3\mathrm{t}}+11(-3{\mathrm{At}}^2-3\mathrm{Bt}+2\mathrm{At}+\mathrm{B}){\mathrm{e}}^{-3\mathrm{t}}-3({\mathrm{At}}^2+\mathrm{Bt}){\mathrm{e}}^{-3\mathrm{t}}=-2{\mathrm{te}}^{-3\mathrm{t}}\\ (-26\mathrm{At}){\mathrm{e}}^{-3\mathrm{t}}+(8\mathrm{A}-13\mathrm{B}){\mathrm{e}}^{-3\mathrm{t}}=-2{\mathrm{te}}^{-3\mathrm{t}}\end{array}$

Step 4: Final conclusion.

Comparing all coefficients of the above equation;

 

$\begin{array}{c}-26\mathrm{A}=-2\\ \mathrm{A}=\frac{1}{13}\\ 8\mathrm{A}-13\mathrm{B}=0\dots \left(3\right)\end{array}$

 

Substitute the value of A in the equation (3),

 

$\begin{array}{c}8\left(\frac{1}{13}\right)-13\mathrm{B}=0\\ \mathrm{B}=\frac{8}{169}\end{array}$

 

Therefore, the particular solution of equation (1),

 ${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)=(\frac{1}{13}{\mathrm{t}}^2+\frac{8}{169}\mathrm{t}){\mathrm{e}}^{-3\mathrm{t}}$