Question

The equation of the curve passing through the origin and satisfying the differetial equation $\frac{dy}{dx}=\sin \left(10x+6y\right)$

• A. $y=\frac{1}{3}{\tan }^{-1}\left(\frac{5\tan 4x}{4-3\tan 4x}\right)-\frac{5x}{3}$
• B. $y=\frac{1}{3}{\tan }^{-1}\left(\frac{5\tan 4x}{4+3\tan 4x}\right)-\frac{5x}{3}$
• C. $y=\frac{1}{3}{\tan }^{-1}\left(\frac{4\tan (4x+ta{n}^{-1}(\frac{3}{4}\left)\right)}{4+3\tan 4x}\right)-\frac{5x}{3}$
• D. None of these

Answer 1

Answer: (D)

None of these

Given:$\frac{dy}{dx}=\sin (10x+6y)$

Let $10x+6y=t$ .......$\left(1\right)$

$\Rightarrow 10+6\frac{dy}{dx}=\frac{dt}{dx}$

$\Rightarrow \frac{dy}{dx}=\frac{1}{6}(\frac{dt}{dx}-10)$

Now, the given differential equation becomes

$\sin t=\frac{1}{6}(\frac{dt}{dx}-10)$

$\Rightarrow 6\sin t=\frac{dt}{dx}-10$

$\Rightarrow \frac{dt}{dx}=6\sin t+10$

$\Rightarrow \frac{dt}{6\sin t+10}=dx$

On integrating both sides,we get

$\Rightarrow \frac{1}{2}\int \frac{dt}{3\sin t+5}=\int dx$ ........$\left(2\right)$

Let $I_1=\int \frac{dt}{3\sin t+5}$

$=\int \frac{dt}{3\left(\frac{2\tan \frac{t}{2}}{1+{\tan }^2\frac{t}{2}}\right)+5}$

$=\int \frac{1+{\tan }^2\frac{t}{2}}{6\tan \frac{t}{2}+5+5{\tan }^2\frac{t}{2}}dt$

Put $u=\tan \frac{t}{2}$

$\Rightarrow \frac{1}{2}{\sec }^2\frac{t}{2}dt=du$

$\Rightarrow dt=\frac{2du}{{\sec }^2\frac{t}{2}}$

$\Rightarrow dt=\frac{2du}{1+{\tan }^2\frac{t}{2}}$

$\Rightarrow dt=\frac{2du}{1+u^2}$

$\therefore I_1=\int \frac{2(1+u^2)du}{(1+u^2)(5u^2+6u+5)}$

$=\frac{2}{5}\int \frac{du}{u^2+\frac{6}{5}u+1}$

$=\frac{2}{5}\int \frac{du}{u^2+2\times \frac{3}{5}u+\frac{9}{25}-\frac{9}{25}+1}$

$=\frac{2}{5}\int \frac{du}{u^2+2\times \frac{3}{5}u+\frac{9}{25}+\frac{16}{25}}$

$=\frac{2}{5}\int \frac{du}{{(u+\frac{3}{5})}^2+{\left(\frac{4}{5}\right)}^2}$

$=\frac{2}{5}\times \frac{5}{4}{\tan }^{-1}\left(\frac{u+\frac{3}{5}}{\frac{4}{5}}\right)$

$=\frac{1}{2}{\tan }^{-1}\left(\frac{5u+3}{4}\right)$

$=\frac{1}{2}{\tan }^{-1}\left(\frac{5\tan \frac{t}{2}+3}{4}\right)$

On putting this in equation $\left(2\right)$,we get

$\frac{1}{4}{\tan }^{-1}\left(\frac{5\tan \frac{t}{2}+3}{4}\right)=x+c$

$\Rightarrow {\tan }^{-1}\left(\frac{5\tan (5x+3y)+3}{4}\right)=4x+4c$

$\Rightarrow {\tan }^{-1}\left(\frac{5\tan (5x+3y)+3}{4}\right)=4x+4c$

$\Rightarrow \frac{5\tan (5x+3y)+3}{4}=\tan (4x+4c)$

$\Rightarrow 5\tan (5x+3y)+3=4\tan (4x+4c)$

When $x=0,y=0$ we get

$\Rightarrow 5\tan 0+3=4\tan 4c$

$\Rightarrow 4c={\tan }^{-1}\frac{3}{4}$

Then,$5\tan (5x+3y)+3=4\tan (4x+{\tan }^{-1}\frac{3}{4})$

$\Rightarrow \tan (5x+3y)=\frac{4}{5}\tan (4x+{\tan }^{-1}\frac{3}{4})-\frac{3}{5}$

$\Rightarrow 5x+3y={\tan }^{-1}[\frac{4}{5}\tan (4x+{\tan }^{-1}\frac{3}{4})-\frac{3}{5}]$

$\Rightarrow 3y={\tan }^{-1}[\frac{4}{5}\tan (4x+{\tan }^{-1}\frac{3}{4})-\frac{3}{5}]-5x$

$\therefore y=\frac{1}{3}{\tan }^{-1}[\frac{4}{5}\tan (4x+{\tan }^{-1}\frac{3}{4})-\frac{3}{5}]-\frac{5x}{3}$