Question

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

• A. $y=\frac{1}{3}ta{n}^{-1}\left(\frac{5tan4x}{4-3tan4x}\right)-\frac{5x}{3}$
• B. $y=\frac{1}{3}ta{n}^{-1}\left(\frac{5tan4x}{4+3tan4x}\right)-\frac{5x}{3}$
• C. $y=\frac{1}{3}ta{n}^{-1}\left(\frac{3+tan4x}{4-3tan4x}\right)-\frac{5x}{3}$
• D. none of these

Answer 1

The correct option is A $y=\frac{1}{3}ta{n}^{-1}\left(\frac{5tan4x}{4-3tan4x}\right)-\frac{5x}{3}$

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

Substituting $10x+6y=t\Rightarrow \frac{dy}{dx}=\frac{1}{6}(\frac{dt}{dx}-10)$

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

Integrating both sides

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

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

$\frac{1}{2}\int \frac{2du}{5u^2+6u+5}=x+c\Rightarrow \frac{1}{5}\int \frac{2du}{u^2+\frac{6}{5}u+1}=x+c\Rightarrow \frac{1}{5}\int \frac{du}{{(u+\frac{3}{5})}^2+{\left(\frac{4}{5}\right)}^2}=x+c\Rightarrow \frac{1}{5}\frac{1}{4/5}{\tan }^{-1}\frac{u+3/5}{4/5}=x+c\Rightarrow \frac{1}{4}{\tan }^{-1}\frac{5u+3}{4}=x+c\Rightarrow 5u+3=4\tan 4(x+c)\Rightarrow 5\tan (5x+3y)+3=4\tan 4(x+c)$

Since it passing through origin

$5\tan 0+3=4\tan 4\left(c\right)\Rightarrow c=\frac{1}{4}{\tan }^{-1}\left(\frac{3}{4}\right)$

Hence

$5\tan (5x+3y)+3=4\tan (4x+{\tan }^{-1}\left(\frac{3}{4}\right))\Rightarrow \tan (5x+3y)=\frac{4}{5}\frac{\tan 4x+\frac{3}{4}}{1-\tan 4x.\frac{3}{4}}-\frac{3}{5}=\frac{4}{5}\frac{(3+4\tan 4x)}{(4-3\tan 4x)}-\frac{3}{5}=\frac{25\tan 4x}{5(4-3\tan 4x)}\Rightarrow y=\frac{1}{3}\left[{\tan }^{-1}\left(\frac{5\tan 4x}{4-3\tan 4x}\right)-5x\right]$