Question

Determine the equation of the curve passing through the origin in the form $y=f(x)$, which satisfies the differential equation $\frac{dy}{dx}=\sin (10x+6y)$

Answer 1

Given, $\frac{dy}{dx}=\sin (10x+6y)$
Let $10x+6y=t....$(i)
$10+6\frac{dy}{dx}=\left(\frac{dt}{dx}\right)$
$\Rightarrow \frac{dy}{dx}=\frac{1}{6}\left(\frac{dt}{dx}-10\right)$
Now, the given differential equation becomes
$\sin t=\frac{1}{6}\left(\frac{dt}{dx}-10\right)$
$\Rightarrow 6\sin t=\frac{dt}{dx}-10$
$\Rightarrow \frac{dt}{dx}=6\sin t+10$
$\Rightarrow \frac{dt}{6\sin t+10}=dx$
On integrating both sides, we get
$\frac{1}{2}\int \frac{dt}{3\sin t+5}=x+c....$(ii)
Let $I_1=\int \frac{dt}{3\sin t+5}=\int \frac{dt}{3\left(\frac{2\tan t/2}{1+{\tan }^{2}t/2}\right)+5}$
$=\int \frac{\left(1+{\tan }^{2}t/2\right)dt}{\left(6\tan \frac{t}{2}+5+5{\tan }^2\frac{t}{2}\right)}$
Put $\tan t/2=u$
$\Rightarrow \frac{1}{2}{\sec }^{2}t/2dt=du\Rightarrow dt=\frac{2du}{{\sec }^{2}t/2}$
$\Rightarrow dt=\frac{2du}{1+{\tan }^{2}t/2}\Rightarrow dt=\frac{2du}{1+u^2}$
$\therefore I_1=\int \frac{2\left(1+u^2\right)du}{\left(1+u^2\right)\left(5u^2+6u+5\right)}=\frac{2}{5}\int \frac{du}{u^2+\frac{6}{5}u+1}$
$=\frac{2}{5}\int \frac{du}{u^2+\frac{6}{5}u+\frac{9}{25}-\frac{9}{25}+1}$
$=\frac{2}{5}\int \frac{du}{{\left(u+\frac{3}{5}\right)}^2+{\left(\frac{4}{5}\right)}^2}=\frac{2}{5}\cdot \frac{5}{4}{\tan }^{-1}\left(\frac{u+3/5}{4/5}\right)$
$=\frac{1}{2}{\tan }^{-1}\left[\frac{5u+3}{4}\right]=\frac{1}{2}{\tan }^{-1}\left[\frac{5\tan t/2+3}{4}\right]$
On putting this in Eq. (ii), we get
$\frac{1}{4}{\tan }^{-1}\left[\frac{5\tan \frac{t}{2}+3}{4}\right]=x+c$
$\Rightarrow {\tan }^{-1}\left[\frac{5\tan \frac{t}{2}+3}{4}\right]=4x+4c$
$\Rightarrow \frac{1}{4}[5\tan (5x+3y)+3]=\tan (4x+4c)$
$\Rightarrow 5\tan (5x+3y)+3=4\tan (4x+4c)$
When $x=0,y=0$, we get
$5\tan 0+3=4\tan (4c)$
$\Rightarrow \frac{3}{4}=\tan 4c\Rightarrow 4c={\tan }^{-1}\frac{3}{4}$
Then, $5\tan (5x+3y)+3=4\tan \left(4x+{\tan }^{-1}\frac{3}{4}\right)$
$\Rightarrow \tan (5x+3y)=\frac{4}{5}\tan \left(4x+\tan \frac{3}{4}\right)-\frac{3}{5}$
$\Rightarrow 5x+3y={\tan }^{-1}\left[\frac{4}{5}\left\{\tan \left(4x+{\tan }^{-1}\frac{3}{4}\right)\right\}-\frac{3}{5}\right]$
$\Rightarrow 3y={\tan }^{-1}\left[\frac{4}{5}\left\{\tan \left(4x+{\tan }^{-1}\frac{3}{4}\right)\right\}-\frac{3}{5}\right]-5x$
$\Rightarrow y=\frac{1}{3}{\tan }^{-1}\left[\frac{4}{5}\left\{\tan \left(4x+{\tan }^{-1}\frac{3}{4}\right)\right\}-\frac{3}{5}\right]-\frac{5x}{3}$