Question

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

Short answer

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

Step-by-step solution

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.

Analyzing the given statement

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

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

Transform the equation (1) in homogeneous equation in u and v

Comparing (1) with the equation,

$$\begin{gathered} \left({\mathrm{a}}_1\mathrm{x}+{\mathrm{b}}_1\mathrm{y}+{\mathrm{c}}_1\right)\mathrm{dx}+\left({\mathrm{a}}_2\mathrm{x}+{\mathrm{b}}_2\mathrm{y}+{\mathrm{c}}_2\right)\mathrm{dy}=0 \\ \mathrm{As}{\mathrm{a}}_1{\mathrm{b}}_2\ne {\mathrm{a}}_2{\mathrm{b}}_1 \end{gathered}$$

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

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

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

Solving (2) and (3),

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

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

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

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

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

which is the required homogeneous equation in u and v.

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),

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

Separating the variables,

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

Integrating both sides,

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

Now, solve the integration part then

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

Now, equation (4) becomes,

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

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

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