Q17E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find a general solution to the differential equation.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{3}\mathbf{y}\mathbf{=}\mathbf{3}{\mathbf{t}}^2\mathbf{-}\mathbf{5}$

Step-by-Step Solution

Answer

The general solution is $\mathbf{y}\mathbf{=}{\mathbf{c}}_1{\mathbf{e}}^{-t}\mathbf{+}{\mathbf{c}}_2{\mathbf{e}}^{3t}\mathbf{-}{\mathbf{t}}^2\mathbf{+}\frac{4}{3}\mathbf{t}\mathbf{+}\frac{1}{9}.$

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

Given that, 

 

The differential equation is,

 

$\mathrm{y}\text{'}\text{'}-2\mathrm{y}\text{'}-3\mathrm{y}=3{\mathrm{t}}^2-5\dots \left(1\right)$

 

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

 

$\mathrm{y}\text{'}\text{'}-2\mathrm{y}\text{'}-3\mathrm{y}=0$

 

The auxiliary equation for the above equation,

 

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

 

Step 2: Find the complementary solution of the given equation.

Solve the auxiliary equation,

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

 

The roots of the auxiliary equation are, 

 

${\mathrm{m}}_1=-1,\&{\mathrm{m}}_2=3$

 

The complementary solution of the given equation is,

 

${\mathrm{y}}_{\mathrm{c}}={\mathrm{c}}_1{\mathrm{e}}^{-\mathrm{t}}+{\mathrm{c}}_2{\mathrm{e}}^{3\mathrm{t}}$

 

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

Assume, the particular solution of equation (1),

 

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

 

Now find the first and second derivatives of the above equation,

 $\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{x}\right)=2\mathrm{At}+\mathrm{B}\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{x}\right)=2\mathrm{A}\end{array}$

 

Substitute the value of ${\mathbf{y}}_p\left(\mathbf{x}\right)\mathbf{,}{\mathbf{y}}_p\mathbf{\text{'}}\left(\mathbf{x}\right)$  and ${\mathbf{y}}_p\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)$ the equation (1),

$\begin{array}{c}\mathrm{y}\text{'}\text{'}-2\mathrm{y}\text{'}-3\mathrm{y}=3{\mathrm{t}}^2-5\\ 2\mathrm{A}-2(2\mathrm{At}+\mathrm{B})-3({\mathrm{At}}^2+\mathrm{Bt}+\mathrm{C})=3{\mathrm{t}}^2-5\\ -3{\mathrm{At}}^2+(-4\mathrm{A}-3\mathrm{B})\mathrm{t}+2\mathrm{A}-2\mathrm{B}-3\mathrm{C}=3{\mathrm{t}}^2-5\end{array}$

Comparing all coefficients of the above equation,

$\begin{array}{c}-3\mathrm{A}=3\Rightarrow \mathrm{A}=-1\\ -4\mathrm{A}-3\mathrm{B}=0\dots \left(3\right)\\ 2\mathrm{A}-2\mathrm{B}-3\mathrm{C}=-5\dots \left(4\right)\end{array}$

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

 $\begin{array}{c}-4(-1)-3\mathrm{B}=0\\ \mathrm{B}=\frac{4}{3}\end{array}$

 

Substitute the value of A and B in the equation (4),

 $\begin{array}{c}2(-1)-2\left(\frac{4}{3}\right)-3\mathrm{C}=-5\\ -3\mathrm{C}=-5+\frac{8}{3}+2\\ \mathrm{c}=\frac{1}{9}\end{array}$

 

 

Substitute the value of A, B, and C in the equation (2),

 

Therefore, the particular solution of equation (1),

$\begin{array}{c}{\mathbf{y}}_p\left(\mathbf{x}\right)\mathbf{=}{\mathbf{At}}^2\mathbf{+}\mathbf{Bt}\mathbf{+}\mathbf{C}\\ {\mathbf{y}}_p\left(\mathbf{x}\right)\mathbf{=}\mathbf{-}{\mathbf{t}}^2\mathbf{+}\frac{4}{3}\mathbf{t}\mathbf{+}\frac{1}{9}\end{array}$

Step 4: Conclusion.

Therefore, the general solution is,

$\begin{array}{c}\mathrm{y}={\mathrm{y}}_{\mathrm{c}}\left(\mathrm{t}\right)+{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)\\ \mathrm{y}={\mathrm{c}}_1{\mathrm{e}}^{-\mathrm{t}}+{\mathrm{c}}_2{\mathrm{e}}^{3\mathrm{t}}-{\mathrm{t}}^2+\frac{4}{3}\mathrm{t}+\frac{1}{9}\end{array}$