Q37E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find general solutions to the nonhomogeneous Cauchy-Euler equations using a variety of parameters.

${\mathbf{t}}^{\mathbf{2}}\mathbf{z}\text{'}\text{'}\mathbf{+}\mathbf{tz}\text{'}\mathbf{+}\mathbf{9}\mathbf{z}\mathbf{=}\mathbf{-}\mathbf{tan}\mathbf{\left(}\mathbf{3}\mathbf{lnt}\mathbf{\right)}$

Step-by-Step Solution

Answer

The general solution of the given equation ${\mathrm{t}}^2\mathrm{z}\text{'}\text{'}+\mathrm{tz}\text{'}+9\mathrm{z}=-\tan \left(3\mathrm{lnt}\right)$ is:

$\mathrm{y}\left(\mathrm{t}\right)={\mathrm{y}}_{\mathrm{h}}\left(\mathrm{t}\right)+{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{c}}_1\cos \left(3\mathrm{lnt}\right)+{\mathrm{c}}_2\sin \left(3\mathrm{lnt}\right)+\frac{\ln \left(\tan \right(3\mathrm{lnt})+\sec (3\mathrm{lnt}\left)\right)}{9}\cos \left(3\mathrm{lnt}\right)$ .

Step 1: Solve the homogeneous equation

First, one needs to solve the homogeneous equation ${\mathrm{t}}^2\mathrm{z}\text{'}\text{'}+\mathrm{tz}\text{'}+9\mathrm{z}=0\cdots \left(1\right)$

The solution of the equation ${\mathrm{at}}^2\mathrm{z}\text{'}\text{'}+\mathrm{btz}\text{'}+\mathrm{cz}=0$ is where t is the root of the equation .

 

From $\left(1\right):\mathrm{a}=\mathrm{b}=1,\mathrm{c}=9$, so we need to solve the quadratic equation ${\mathrm{r}}^2+9=0$ which has solutions ${\mathrm{r}}_{1,2}=\pm 3\mathrm{i}$. Since ${\mathrm{t}}^{3\mathrm{i}}={\mathrm{e}}^{3\mathrm{ilnt}}=\cos \left(3\mathrm{lnt}\right)+\mathrm{isin}\left(3\mathrm{lnt}\right)$

solutions of (1) are ${\mathrm{y}}_1\left(\mathrm{t}\right)=\cos \left(3\mathrm{lnt}\right), {\mathrm{y}}_2\left(\mathrm{t}\right)=\sin \left(3\mathrm{lnt}\right)$

 

To find a particular solution to the given equation we need to divide both sides by ${\mathrm{t}}^2$

$\mathrm{z}\text{'}\text{'}+\frac{1}{\mathrm{t}}\mathrm{z}\text{'}+\frac{9}{{\mathrm{t}}^2}\mathrm{z}=-\frac{\tan \left(3\mathrm{lnt}\right)}{{\mathrm{t}}^2}$

Step 2: Solving v 1

By Variation of Parameters method, the particular solution has the form:

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

Where,

$\begin{array}{rcl}& & {\mathrm{v}}_1\left(\mathrm{t}\right)=\int \frac{-\mathrm{g}\left(\mathrm{t}\right){\mathrm{y}}_2\left(\mathrm{t}\right)}{{\mathrm{y}}_1{\left(\mathrm{t}\right)\mathrm{y}}_2^{}{\text{'}\left(\mathrm{t}\right)-\mathrm{y}}_1^{}\text{'}\left(\mathrm{t}\right){\mathrm{y}}_2\left(\mathrm{t}\right)}\mathrm{dt}\\ & =& \int \frac{\frac{\tan \left(3\mathrm{lnt}\right)}{{\mathrm{t}}^2}\sin \left(3\mathrm{lnt}\right)}{\cos \left(3\mathrm{lnt}\right)\times \cos \left(3\mathrm{lnt}\right)\times 3\frac{1}{\mathrm{t}}+\sin \left(3\mathrm{lnt}\right)\times \sin \left(3\mathrm{lnt}\right)\times 3\frac{1}{\mathrm{t}}}\mathrm{dt}\\ & =& \int \frac{{\sin }^2\left(3\mathrm{lnt}\right)}{\frac{3}{\mathrm{t}}\left({\cos }^2\left(3\mathrm{lnt}\right)+\mathrm{si}{\mathrm{n}}^2\left(3\mathrm{lnt}\right)\right)}\mathrm{dt}\\ & =& \frac{1}{9}\int \frac{{\sin }^2\mathrm{x}}{\mathrm{cosx}}\mathrm{dx}=\frac{1}{9}\int \frac{1-{\cos }^2\mathrm{x}}{{\cos }^2\mathrm{x}}\mathrm{dx}\\ & =& \frac{1}{9}\left(\int \frac{\mathrm{dx}}{\mathrm{cosx}}-\mathrm{cosxdx}\right)=\frac{\ln (\mathrm{tanx}+\mathrm{secx})-\mathrm{sinx}}{9}+{\mathrm{C}}_1\\ & =& \frac{\ln \left(\tan \right(3\mathrm{lnt})+\sec (3\mathrm{lnt}\left)\right)-\sin \left(3\mathrm{lnt}\right)}{9}+{\mathrm{C}}_1\end{array}$

Here $\mathrm{x}=3\mathrm{lnt}\Rightarrow \mathrm{dx}=\frac{3}{\mathrm{t}}\mathrm{dt}$

Step 3: Solving v 2

Solve the other part,

$\begin{array}{rcl}& & {\mathrm{v}}_2\left(\mathrm{t}\right)=\int \frac{\mathrm{g}\left(\mathrm{t}\right){\mathrm{y}}_1\left(\mathrm{t}\right)}{{\mathrm{y}}_1{\left(\mathrm{t}\right)\mathrm{y}}_2^{}{\text{'}\left(\mathrm{t}\right)-\mathrm{y}}_1^{}\text{'}\left(\mathrm{t}\right){\mathrm{y}}_2\left(\mathrm{t}\right)}\mathrm{dt}\\ & =& \int \frac{-\frac{\tan \left(3\mathrm{lnt}\right)}{{\mathrm{t}}^2}\cos \left(3\mathrm{lnt}\right)}{\cos \left(3\mathrm{lnt}\right)\times \cos \left(3\mathrm{lnt}\right)\times 3\frac{1}{\mathrm{t}}+\sin \left(3\mathrm{lnt}\right)\times \sin \left(3\mathrm{lnt}\right)\times 3\frac{1}{\mathrm{t}}}\mathrm{dt}\\ & =& \int \frac{\frac{\sin \left(3\mathrm{lnt}\right)}{{\mathrm{t}}^2\cos \left(3\mathrm{lnt}\right)}\cos \left(3\mathrm{lnt}\right)}{\frac{3}{\mathrm{t}}\left({\cos }^2\left(3\mathrm{lnt}\right)+\mathrm{si}{\mathrm{n}}^2\left(3\mathrm{lnt}\right)\right)}\mathrm{dt}\\ & =& -\frac{1}{9}\int \mathrm{sinxdx}=\frac{1}{9}\mathrm{cosx}+{\mathrm{C}}_2\\ & =& \frac{\cos \left(3\mathrm{lnt}\right)}{9}+{\mathrm{C}}_2\end{array}$

Here $\mathrm{x}=3\mathrm{lnt}\Rightarrow \mathrm{dx}=\frac{3}{\mathrm{t}}\mathrm{dt}$

Step 4: Substitute the values

Hence,

$$\begin{gathered} {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\frac{\ln \left(\tan \right(3\mathrm{lnt})+\sec (3\mathrm{lnt}\left)\right)-\sin \left(3\mathrm{lnt}\right)}{9}\cos \left(3\mathrm{lnt}\right)+\frac{\cos \left(3\mathrm{lnt}\right)}{9}\sin \left(3\mathrm{lnt}\right) \\ =\frac{\ln \left(\tan \right(3\mathrm{lnt})+\sec (3\mathrm{lnt}\left)\right)}{9}\cos \left(3\mathrm{lnt}\right) \end{gathered}$$

 

The general solution is:

$\mathrm{y}\left(\mathrm{t}\right)={\mathrm{y}}_{\mathrm{h}}\left(\mathrm{t}\right)+{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{c}}_1\cos \left(3\mathrm{lnt}\right)+{\mathrm{c}}_2\sin \left(3\mathrm{lnt}\right)+\frac{\ln \left(\tan \right(3\mathrm{lnt})+\sec (3\mathrm{lnt}\left)\right)}{9}\cos \left(3\mathrm{lnt}\right)$.