Q16E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 11–18, find a general solution to the differential equation.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{6}\mathbf{y}\mathbf{=}\mathbf{18}{\mathbf{t}}^{\mathbf{2}}$

Step-by-Step Solution

Answer

The general solution is $y\left(t\right)=c_1{e}^{-3t}+c_2{e}^{-2t}+3t^2-5t+\frac{19}{6}$.

Step 1: Find a particular solution

The auxiliary equation is $r^2+5r+6=0$.

 

Two independent solutions are $r=-2,-3$.

 

Then $y_1=e^{-3t},v_2=e^{-2t}$

$y_h\left(t\right)=c_1{e}^{-3t}+c_2{e}^{-2t}$

 

The particular solution is ${\mathrm{y}}_{\mathrm{p}}={\mathrm{v}}_1\left(\mathrm{t}\right){\mathrm{e}}^{-3\mathrm{t}}+{\mathrm{v}}_2\left(\mathrm{t}\right){\mathrm{e}}^{-2\mathrm{t}}$.

Step 2: Find v 1 ' and v 1

 Now,

$\begin{array}{rcl}{v}_1\text{'}& =& \frac{-f\left(t\right)y_2\left(t\right)}{a\left[y_1\left(t\right)y{\text{'}}_2\left(t\right)-y{\text{'}}_1\left(t\right)y_2\left(t\right)\right]}\\ & =& \frac{-18t^2.e^{-2t}}{\left[e^{-3t}\left(-2e^{-2t}\right)+3e^{-3t}{e}^{-2t}\right]}\\ & =& \frac{-18t^2.e^{-2t}}{\left[-2e^{-5t}+3e^{-5t}\right]}\\ & =& \frac{-18t^2.e^{-2t}}{\left[e^{-5t}\right]}\\ & =& -18t^2{e}^{3t}\end{array}$

 

Now integrating this by integration by parts formula

$\begin{array}{rcl}{v}_1\left(t\right)& =& \int -18t^2{e}^{3t}dt\\ & =& \frac{-2e^{3t}}{3}\left(2-6t+9t^2\right)+C\end{array}$

Step 3: Determine v 2 ' and v 2

$\begin{array}{rcl}{v}_2\text{'}& =& \frac{f\left(t\right)y_1\left(t\right)}{a\left[y_1\left(t\right)y{\text{'}}_2\left(t\right)-y{\text{'}}_1\left(t\right)y_2\left(t\right)\right]}\\ & =& \frac{18t^2.e^{-3t}}{\left[e^{-3t}\left(-2e^{-2t}\right)+3e^{-3t}{e}^{-2t}\right]}\\ & =& 18t^2{e}^{2t}\end{array}$

Integrate this by integration by parts.

$\begin{array}{rcl}{y}_2\left(t\right)& =& \int 18t^2{e}^{2t}dt\\ & =& \frac{9}{2}{e}^{2t}\left(1-2t+2t^2\right)+C\end{array}$ 

Thus, a particular solution is:

$\begin{array}{rcl}{y}_p& =& \left(\frac{-2e^{3t}}{3}\left(2-6t+9t^2\right)+C\right)e^{-3t}+\left(\frac{9}{2}{e}^{2t}\left(1-2t+2t^2\right)+C\right)e^{-2t}\\ y_p& =& 3t^2-5t+\frac{19}{6}\end{array}$ 

 

Therefore, the general solution is:

 $$\begin{gathered} y\left(t\right)=y_h\left(t\right)+y_p\left(t\right) \\ y\left(t\right)=c_1{e}^{-3t}+c_2{e}^{-2t}+3t^2-5t+\frac{19}{6} \end{gathered}$$