Question

The equation of the curve passing through origin in the form $y=f\left(x\right)$, satisfing the differential coefficient by the relation ${\sin }^{-1}\left(\frac{dy}{dx}\right)=10x+6y$ is:

• A. $\frac{1}{3}{\tan }^{-1}\left(\frac{4\tan 4x}{5-3\tan 4x}\right)+\frac{4}{3}x$
• B. $\frac{1}{3}{\tan }^{-1}\left(\frac{5\tan 4x}{4-3\tan 4x}\right)-\frac{5}{3}x$
• C. $\frac{1}{3}\left({\tan }^{-1}\left(\frac{4\tan 4x}{5-3\tan 4x}\right)\right)-\frac{4}{3}x$
• D. $\frac{1}{3}{\tan }^{-1}\left(\frac{5\tan 4x}{4-3\tan 4x}\right)+\frac{5}{3}x$

Answer 1

The correct option is B $\frac{1}{3}{\tan }^{-1}\left(\frac{5\tan 4x}{4-3\tan 4x}\right)-\frac{5}{3}x$

Given differential equation is $\frac{dy}{dx}=\sin (10x+6y)$,

Let $10x+6y=t$

$⟹\frac{dy}{dx}=\frac{\frac{dt}{dx}-10}{6}$

$⟹\int \frac{1}{6\sin t+10}dt=\int dx$

$⟹\int \frac{{\sec }^2\frac{t}{2}}{12\tan \frac{t}{2}+10{\tan }^2\frac{t}{2}+10}dt=\int dx$

We can solve the above integral by taking $\tan \frac{t}{2}=k⟹{\frac{1}{2}\sec }^2\frac{t}{2}dt=dk$

$⟹\int \frac{1}{5k^2+6k+5}dk=\int dx⟹\int \frac{1}{{(\sqrt{5}k+\frac{3}{\sqrt{5}})}^2+\frac{16}{5}}dk=x⟹\frac{1}{4}{\tan }^{-1}\left(\frac{5k+3}{4}\right)=x⟹\frac{1}{4}{\tan }^{-1}\left(\frac{5\tan (5x+3y)+3}{4}\right)=x$

Simplyfying gives,

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