Q2E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 1–8, find a general solution to the differential equation using the method of variation of parameters. $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{sect}$

Step-by-Step Solution

Answer

The general solution is  $\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{c}}_1\mathbf{cost}\mathbf{+}{\mathbf{c}}_2\mathbf{sint}\mathbf{+}\left(\mathbf{ln}\left|\mathbf{cost}\right|\right)\mathbf{cost}\mathbf{+}\mathbf{tsint}$. 

Step 1: Find a particular solution.

The homogenous equation is ${\mathrm{r}}^2+1=0$.

 

Two independent solutions are $\mathrm{r}=\pm \mathrm{i}$.

 

${\mathrm{y}}_1=\mathrm{cost},{\mathrm{y}}_2=\mathrm{sint}$

 

Then ${\mathrm{y}}_{\mathrm{h}}\left(\mathrm{t}\right)={\mathrm{c}}_1\mathrm{cost}+{\mathrm{c}}_2\mathrm{sint}$

 

The particular solution is  ${\mathrm{y}}_{\mathrm{p}}={\mathrm{v}}_1\left(\mathrm{t}\right)\mathrm{cost}+{\mathrm{v}}_2\left(\mathrm{t}\right)\mathrm{sint}$

Step 2: evaluate v 1 and v 2 .

Here ${\mathrm{y}}_{\mathrm{p}}={\mathrm{v}}_1\left(\mathrm{t}\right)\mathrm{cost}+{\mathrm{v}}_2\left(\mathrm{t}\right)\mathrm{sint}$

 

And referring to (9) and solve the system then

  $\begin{array}{c}{\mathrm{v}}_1\text{'}\mathrm{cost}+{\mathrm{v}}_2\text{'}\mathrm{sint}=0\\ -{\mathrm{v}}_1\text{'}\mathrm{sint}+{\mathrm{v}}_2\text{'}\mathrm{cost}=\frac{\mathrm{f}}{\mathrm{a}}\\ -{\mathrm{v}}_1\text{'}\mathrm{sint}+{\mathrm{v}}_2\text{'}\mathrm{cost}=\mathrm{sect}\end{array}$

Step 3: Find v 1 ' and v 1 .

$\begin{array}{c}{\mathrm{v}}_1\text{'}=\frac{-\mathrm{f}\left(\mathrm{t}\right){\mathrm{y}}_2\left(\mathrm{t}\right)}{\mathrm{a}\left[{\mathrm{y}}_1\left(\mathrm{t}\right)\mathrm{y}{\text{'}}_2\left(\mathrm{t}\right)-\mathrm{y}{\text{'}}_1\left(\mathrm{t}\right){\mathrm{y}}_2\left(\mathrm{t}\right)\right]}\\ =\frac{-\mathrm{sect}.\mathrm{sint}}{1[-{\cos }^2\mathrm{t}.-{\sin }^2\mathrm{t}]}\\ =\frac{-\mathrm{sect}.\mathrm{sint}}{[-{\cos }^2\mathrm{t}-{\sin }^2\mathrm{t}]}\\ =\mathrm{tant}\end{array}$

Now integrating this.

$\begin{array}{l}{\mathrm{v}}_1\left(\mathrm{t}\right)=\int -\mathrm{tant}\mathrm{dt}\\ =\ln \left|\mathrm{cost}\right|+\mathrm{C}\end{array}$

Step 4: Determine v 2 ' and v 2 .

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

Integrate this.

$\begin{array}{l}{\mathrm{v}}_2\left(\mathrm{t}\right)=\int 1\mathrm{dt}\\ =\mathrm{t}+\mathrm{C}\end{array}$

Thus particular solution is

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}=(\ln \left|\mathrm{cost}\right|+\mathrm{C})\mathrm{cost}+(\mathrm{t}+\mathrm{C})\mathrm{sint}\\ {\mathrm{y}}_{\mathrm{p}}=\left(\ln \left|\mathrm{cost}\right|\right)\mathrm{cost}+\mathrm{tsint}\end{array}$

Thus, general solution is:

$\begin{array}{c}\mathrm{y}\left(\mathrm{t}\right)={\mathrm{y}}_{\mathrm{h}}\left(\mathrm{t}\right)+{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)\\ \mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1\mathrm{cost}+{\mathrm{c}}_2\mathrm{sint}+\left(\ln \left|\mathrm{cost}\right|\right)\mathrm{cost}+\mathrm{tsint}\end{array}$