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