Q11E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{tant}\mathbf{+}{\mathbf{e}}^{\mathbf{3}\mathbf{t}}\mathbf{-}\mathbf{1}$

Step-by-Step Solution

Answer

The general solution is $\text{y}\left(t\right)=c_{1}cost+c_{2}sint-ln\left|sect+tant\right|.cost+\frac{1}{10}{e}^{3t}-1$.

Step 1: Find a particular solution .

The homogenous equation is $r^2+1=0$.

 

Two independent solutions are $r=\pm i$.

 

Then $y_1=\cos t,y_2=\sin t$

$y_h\left(t\right)=c_1\cos t+c_2\sin t$

The particular solution is $y_p=v_1\left(t\right)cost+v_2\left(t\right)sint$

Step 2: Evaluate v 1     and     v 2

Here $y_p=v_1\left(t\right)\cos t+v_2\left(t\right)\sin t$

And referring to (9) $y\left(t\right)=c_1{e}^{\alpha t}\cos \beta t c_2{e}^{\alpha t}\sin \beta t$ and solve the system by taking derivatives then;

 

Here $f=\tan t+e^{3t}-1$

$\begin{array}{rcl}{v}_1\text{'}\cos t+v_2\text{'}\sin t& =& 0\\ -v_1\text{'}\sin t+v_2\cos t& =& \frac{f}{a}\\ -v_1\text{'}\sin t+v_2\cos t& =& \tan t+e^{3t}-1\end{array}$

Step 3: Find v 1 ' and v 1

$\begin{array}{rcl}{v}_1\text{'}& =& \frac{-f\left(t\right)y_2\left(t\right)}{a\left[y_1\left(t\right)y{\text{'}}_2\left(t\right)-y{\text{'}}_1\left(t\right)y_2\left(t\right)\right]}\\ & =& \frac{-\left(\tan t+e^{3t}-1\right).\sin t}{\left[{\cos }^{2}t+{\sin }^{2}t\right]}\\ & =& -\left(\tan t+e^{3t}-1\right).\sin t\end{array}$

Now integrating this.

$\begin{array}{rcl}{v}_1\left(t\right)& =& \int -\left(\tan t+e^{3t}-1\right).\sin tdt\\ & =& \sin t-\ln \left|\sec t+\tan t\right|+\frac{1}{10}{e}^{3t}\left(\cos t-3\sin t\right)-\cos t+C\end{array}$

Step 4: Determine v 2 ' and v 2

$\begin{array}{rcl}{v}_2\text{'}& =& \frac{f\left(t\right)y_1\left(t\right)}{a\left[y_1\left(t\right)y{\text{'}}_2\left(t\right)-y{\text{'}}_1\left(t\right)y_2\left(t\right)\right]}\\ & =& \frac{\left(\tan t+e^{3t}-1\right).\cos t}{\left[{\cos }^{2}t+{\sin }^{2}t\right]}\\ & =& \left(\tan t+e^{3t}-1\right).\cos t\end{array}$

Integrate this.

$\begin{array}{rcl}{v}_2\left(t\right)& =& \int -\frac{1}{3}{e}^{3t}dt\\ & =& -\cos t+\frac{1}{10}{e}^{3t}\left(\cos t-3\sin t\right)-\cos t+C\end{array}$

Thus, the particular solution is:

$\begin{array}{rcl}{y}_p& =& \sin t-\ln \left|\sec t+\tan t\right|+\frac{1}{10}{e}^{3t}\left(\cos t-3\sin t\right)-\cos t+C-\cos t+\frac{1}{10}{e}^{3t}\left(\cos t-3\sin t\right)-\cos t+C\\ y_p& =& \sin t-\ln \left|\sec t+\tan t\right|+\frac{1}{10}{e}^{3t}\left(\cos t-3\sin t\right)-\cos t-\cos t+\frac{1}{10}{e}^{3t}\left(\cos t-3\sin t\right)-\cos t\\ y_p& =& -\ln \left|\sec t+\tan t\right|.\cos t+\frac{1}{10}{e}^{3t}-1\end{array}$

Therefore, the general solution is:

$$\begin{gathered} \mathrm{y}\left(\mathrm{t}\right)={\mathrm{y}}_{\mathrm{h}}\left(\mathrm{t}\right)+{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right) \\ \mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1\mathrm{cost}+{\mathrm{c}}_2\mathrm{sint}-\ln \left|\mathrm{sect}+\mathrm{tant}\right|.\mathrm{cost}+\frac{1}{10}{\mathrm{e}}^{3\mathrm{t}}-1 \end{gathered}$$