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Q 2.6-36E

Expert-verified
Found in: Page 77

### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

# Question: In Problems 33–40, solve the equation given in: Problem 4.

The solution of the given equation is $\frac{-2}{\sqrt{3}}{\mathbf{tan}}^{-1}\left(\frac{\sqrt{3}\mathbf{\left(}\mathbf{t}\mathbf{-}\mathbf{1}\mathbf{\right)}}{\left(x+3\right)}\right)\mathbf{+}\mathbf{ln}\left|\mathbf{\left(}\mathbf{3}{\left(t-1\right)}^{2}\mathbf{+}\mathbf{\left(}\mathbf{x}\mathbf{+}\mathbf{3}{\mathbf{\right)}}^{2}\right|\mathbf{=}\mathbf{C}$

See the step by step solution

## Step 1: Concept and definition

Homogeneous differential equation is the equation of the form $\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)\mathrm{dy}=\mathrm{g}\left(\mathrm{x},\mathrm{y}\right)\mathrm{dx}$ , where the degree of $\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)$ and $\mathrm{g}\left(\mathrm{x},\mathrm{y}\right)$ is same.

## Step 2: Analyzing the given statement

One has to solve the equation given in problem 4, i.e.

$\left(t+x+2\right)dx+\left(3t-x-6\right)dt=0......\left(1\right)$

## Step 3: Transform the equation (1) in homogeneous equation in u and v

Comparing (1) with the equation,

$\left({\mathrm{a}}_{1}\mathrm{x}+{\mathrm{b}}_{1}\mathrm{y}+{\mathrm{c}}_{1}\right)\mathrm{dx}+\left({\mathrm{a}}_{2}\mathrm{x}+{\mathrm{b}}_{2}\mathrm{y}+{\mathrm{c}}_{2}\right)\mathrm{dy}=0\phantom{\rule{0ex}{0ex}}\mathrm{As}{\mathrm{a}}_{1}{\mathrm{b}}_{2}\ne {\mathrm{a}}_{2}{\mathrm{b}}_{1}$

Here, take x as the dependent variable and t as the independent variable.

Let $t=u+handx=v+k$ , where h and k satisfy,

$\begin{array}{l}h+k+2=0......\left(2\right)\\ 3h-k-6=0......\left(3\right)\end{array}$

Solving (2) and (3),

role="math" localid="1663929399330" $\mathrm{h}=1\text{\hspace{0.17em}}\mathrm{and}\text{\hspace{0.17em}}\mathrm{k}=-3$

$\mathrm{Now}\mathrm{as}\mathrm{t}=\mathrm{u}+\mathrm{h}\mathrm{and}\mathrm{x}=\mathrm{v}+\mathrm{k}$

$\mathrm{Therefore},dt=du\mathrm{and}dx=dv$

Substituting these values of x, t, dx, dt in equation (1),

$\begin{array}{c}\left(u+1+v-3+2\right)dx+\left(3\left(u+1\right)-\left(v-3\right)-6\right)dt=0\\ \left(u+v\right)dv+\left(3u-v\right)du=0\\ \frac{dv}{du}=\frac{-\left(u+v\right)}{3u-v}\\ \frac{dv}{du}=\frac{\left(\frac{v}{u}+1\right)}{1-3\left(\frac{v}{u}\right)}......\left(4\right)\end{array}$

which is the required homogeneous equation in u and v.

## Step 4: Find solution of equation (4)

Let $\frac{v}{u}\mathbf{=}\mathbf{z}$ in equation (4)

Differentiating both sides,

$\begin{array}{c}dz=\frac{udu-vdv}{{u}^{2}}\\ vdz=du-\frac{v}{u}dv\\ vdz=du-zdv\\ \frac{dv}{du}=u\frac{dz}{dv}+z\end{array}$

Substitute this value of localid="1663929574069" $\frac{\mathrm{dv}}{\mathrm{du}}$ in equation (4),

$\begin{array}{c}u\frac{dz}{dv}+z=\frac{\left(z+1\right)}{1-3z}\\ u\frac{dz}{dv}=\frac{\left(z+1\right)}{1-3z}-z\\ u\frac{dz}{dv}=\frac{z+1+3{z}^{2}-z}{1-3z}\\ u\frac{dz}{dv}=\frac{3{z}^{2}+1}{1-3z}\end{array}$

Separating the variables,

$\frac{3z-1}{-3{z}^{2}-1}dz=\frac{1}{v}dv$

Integrating both sides,

$\begin{array}{c}\int \frac{1-3z}{3{z}^{2}+1}dz=\int \frac{1}{u}du\\ \int \left(\frac{1}{3{z}^{2}+1}-\frac{3z}{3{z}^{2}+1}\right)dz=\int \frac{1}{u}du\end{array}$

Now, solve the integration part then

$\begin{array}{c}\int \frac{1}{3{z}^{2}+1}dz=\frac{1}{\sqrt{3}}{\mathrm{tan}}^{-1}\sqrt{3}z\\ 3\int \frac{z}{3{z}^{2}+1}dz=-\frac{1}{2}\mathrm{ln}\left|3{z}^{2}+1\right|\end{array}$

Now, equation (4) becomes,

$\begin{array}{c}\frac{1}{\sqrt{3}}{\mathrm{tan}}^{-1}\sqrt{3}z-\frac{1}{2}\mathrm{ln}\left|3{z}^{2}+1\right|=\mathrm{ln}u+C\\ \frac{-2}{\sqrt{3}}{\mathrm{tan}}^{-1}\sqrt{3}z+\mathrm{ln}\left|3{z}^{2}+1\right|=-2\mathrm{ln}u+C\\ \frac{-2}{\sqrt{3}}{\mathrm{tan}}^{-1}\sqrt{3}z+\mathrm{ln}\left|3{z}^{2}+1\right|=-\mathrm{ln}\left|{u}^{2}\right|+C\\ \frac{-2}{\sqrt{3}}{\mathrm{tan}}^{-1}\sqrt{3}z+\mathrm{ln}\left|3{z}^{2}+1\right|+\mathrm{ln}\left|{u}^{2}\right|=C\end{array}$

$\begin{array}{c}\frac{-2}{\sqrt{3}}{\mathrm{tan}}^{-1}\sqrt{3}z+\mathrm{ln}\left|\left(3{z}^{2}+1\right){u}^{2}\right|=C\\ \frac{-2}{\sqrt{3}}{\mathrm{tan}}^{-1}\sqrt{3}\left(\frac{v}{u}\right)+\mathrm{ln}\left|\left(3{\left(\frac{v}{u}\right)}^{2}+1\right){u}^{2}\right|=C\\ \frac{-2}{\sqrt{3}}{\mathrm{tan}}^{-1}\sqrt{3}\left(\frac{t-1}{x+3}\right)+\mathrm{ln}\left|\left(3{\left(\frac{t-1}{x+3}\right)}^{2}+1\right){\left(x+3\right)}^{2}\right|=C\\ \frac{-2}{\sqrt{3}}{\mathrm{tan}}^{-1}\sqrt{3}\left(\frac{t-1}{x+3}\right)+\mathrm{ln}\left|\left(3\frac{{\left(t-1\right)}^{2}+{\left(x+3\right)}^{2}}{{\left(x+3\right)}^{2}}\right){\left(x+3\right)}^{2}\right|=C\\ -\frac{2}{\sqrt{3}}{\mathrm{tan}}^{-1}\left(\frac{\sqrt{3}\left(t-1}{\left(x+3\right)}\right)+\mathrm{ln}\left|3{\left(t-1\right)}^{2}+{\left(x+3\right)}^{2}\right|=C\end{array}$

Hence, the solution of given equation is $\frac{-2}{\sqrt{3}}{{\mathbf{tan}}}^{-1}\left(\frac{\sqrt{3}\mathbf{\left(}\mathbf{t}\mathbf{-}\mathbf{1}\mathbf{\right)}}{\left(x+3\right)}\right){\mathbf{+}}{\mathbf{ln}}\left|\mathbf{\left(}\mathbf{3}{\left(t-1\right)}^{2}\mathbf{+}\mathbf{\left(}\mathbf{x}\mathbf{+}\mathbf{3}{\mathbf{\right)}}^{2}\right|{\mathbf{=}}{\mathbf{C}}$.