Q22E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 22 through 25, use a variation of parameters to find a general solution to the differential equation given that the functions and are linearly independent solutions to the corresponding homogeneous equation for t> 0.${\mathbf{t}}^2\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{4}\mathbf{ty}\mathbf{\text{'}}\mathbf{+}\mathbf{6}\mathbf{y}\mathbf{=}{\mathbf{t}}^3\mathbf{+}\mathbf{1}\mathbf{;}{\mathbf{y}}_1\mathbf{=}{\mathbf{t}}^2\mathbf{,}{\mathbf{y}}_2\mathbf{=}{\mathbf{t}}^3$

Step-by-Step Solution

Answer

The general solution is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{t}}^2+{\mathrm{c}}_2{\mathrm{t}}^3+\left(-\mathrm{t}+\frac{{\mathrm{t}}^{-2}}{2}\right){\mathrm{t}}^2+\left(\ln \left|\mathrm{t}\right|-\frac{{\mathrm{t}}^{-3}}{3}\right){\mathrm{t}}^3$.

Step 1:Find a particular solution

Given the differential equation is ${\mathrm{t}}^2\mathrm{y}\text{'}\text{'}-4\mathrm{ty}\text{'}+6\mathrm{y}={\mathrm{t}}^3+1.$

 

And ${\mathrm{y}}_1={\mathrm{t}}^2,{\mathrm{y}}_2={\mathrm{t}}^3$

 

${\mathrm{y}}_{\mathrm{h}}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{t}}^2+{\mathrm{c}}_2{\mathrm{t}}^3$

 

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

Step 2: Evaluate v 1    and    v 2

Here ${\mathrm{y}}_{\mathrm{p}}={\mathrm{v}}_1\left(\mathrm{t}\right){\mathrm{t}}^2+{\mathrm{v}}_2\left(\mathrm{t}\right){\mathrm{t}}^3$

$\begin{array}{c}{\mathrm{v}}_1\text{'}=\frac{-\mathrm{f}\left(\mathrm{t}\right){\mathrm{y}}_2\left(\mathrm{t}\right)}{\mathrm{a}\left[{\mathrm{y}}_1\left(\mathrm{t}\right)\mathrm{y}{\text{'}}_2\left(\mathrm{t}\right)-\mathrm{y}{\text{'}}_1\left(\mathrm{t}\right){\mathrm{y}}_2\left(\mathrm{t}\right)\right]}\\ =\frac{-({\mathrm{t}}^3+1){\mathrm{t}}^3}{\left[{\mathrm{t}}^2({\mathrm{t}}^2.3{\mathrm{t}}^2-2\mathrm{t}.{\mathrm{t}}^3\right]}\\ =\frac{-({\mathrm{t}}^3+1){\mathrm{t}}^3}{\left[{\mathrm{t}}^6\right]}\end{array}$

Now integrate the above result.

 $\begin{array}{c}{\mathrm{v}}_1\left(\mathrm{t}\right)=\int \frac{-({\mathrm{t}}^3+1){\mathrm{t}}^3}{\left[{\mathrm{t}}^6\right]}\mathrm{dt}\\ =-\mathrm{t}+\frac{{\mathrm{t}}^{-2}}{2}\end{array}$

Step 3: Determine v 2 ' and v 2

$\begin{array}{c}{\mathrm{v}}_2\text{'}=\frac{\mathrm{f}\left(\mathrm{t}\right){\mathrm{y}}_1\left(\mathrm{t}\right)}{\mathrm{a}\left[{\mathrm{y}}_1\left(\mathrm{t}\right)\mathrm{y}{\text{'}}_2\left(\mathrm{t}\right)-\mathrm{y}{\text{'}}_1\left(\mathrm{t}\right){\mathrm{y}}_2\left(\mathrm{t}\right)\right]}\\ =\frac{({\mathrm{t}}^3+1){\mathrm{t}}^2}{\left[{\mathrm{t}}^2({\mathrm{t}}^2.3{\mathrm{t}}^2-2\mathrm{t}.{\mathrm{t}}^3\right]}\\ =\frac{({\mathrm{t}}^3+1){\mathrm{t}}^2}{\left[{\mathrm{t}}^6\right]}\\ =\frac{({\mathrm{t}}^6+{\mathrm{t}}^3)}{\left[{\mathrm{t}}^6\right]}\end{array}$

Integrate the above equation.

$\begin{array}{c}{\mathrm{v}}_2\left(\mathrm{t}\right)=\int \frac{({\mathrm{t}}^6+{\mathrm{t}}^3)}{\left[{\mathrm{t}}^6\right]}\mathrm{dt}\\ =\ln \left|\mathrm{t}\right|-\frac{{\mathrm{t}}^{-3}}{3}\end{array}$

Thus the particular solution is:

  ${\mathrm{y}}_{\mathrm{p}}=\left(-\mathrm{t}+\frac{{\mathrm{t}}^{-2}}{2}\right){\mathrm{t}}^2+\left(\ln \left|\mathrm{t}\right|-\frac{{\mathrm{t}}^{-3}}{3}\right){\mathrm{t}}^3$

And the general solution is:

$\begin{array}{c}\mathrm{y}\left(\mathrm{t}\right)={\mathrm{y}}_{\mathrm{h}}\left(\mathrm{t}\right)+{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)\\ \mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{t}}^2+{\mathrm{c}}_2{\mathrm{t}}^3+\left(-\mathrm{t}+\frac{{\mathrm{t}}^{-2}}{2}\right){\mathrm{t}}^2+\left(\ln \left|\mathrm{t}\right|-\frac{{\mathrm{t}}^{-3}}{3}\right){\mathrm{t}}^3\end{array}$