Question

The solution of the differential equation $\frac{dy}{dx}=\sin (10x+6y)$ is

• A. $5\tan (5x-3y)=4\tan (4x+k)+3$
• B. $5\tan (5x+3y)=4\tan (4x+k)-3$
• C. $5\tan (5x-3y)=4\tan (4x+k)-3$
• D. None of these

Answer 1

The correct option is B $5\tan (5x+3y)=4\tan (4x+k)-3$

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

Intregrating both side
$\int \frac{dt}{6\sin t+10}=\int dx+c\Rightarrow \int \frac{dt}{12\sin \frac{t}{2}\cos \frac{t}{2}+10}=\int dx+c$
Multiplying and dividing by ${\sec }^2\frac{t}{2}$
$\Rightarrow \int \frac{{\sec }^2\frac{t}{2}}{12\tan \frac{t}{2}+10{\sec }^2\frac{t}{2}}dt=\int dx+c$
Taking
$u=\tan \frac{t}{2}$
$\Rightarrow du=\frac{1}{2}{\sec }^2\frac{t}{2}dt\Rightarrow dt=\frac{2du}{1+u^2}$

$\Rightarrow \int \frac{2du}{12u+10(1+u^2)}=x+c$
$\Rightarrow \int \frac{du}{5u^2+6u+5}=x+c$
$\Rightarrow \int \frac{du}{{(\sqrt{5}u+\frac{3}{\sqrt{5}})}^2+\frac{16}{5}}=x+c$

Taking
$(\sqrt{5}u+\frac{3}{\sqrt{5}})=s\Rightarrow ds=\sqrt{5}du$

$\Rightarrow \frac{1}{\sqrt{5}}\int \frac{ds}{s^2+{\left(\frac{4}{\sqrt{5}}\right)}^2}=x+c$
$\Rightarrow \frac{1}{4}{\tan }^{-1}\frac{\sqrt{5}s}{4}=x+c$
$\Rightarrow 5u+3=4\tan (4x+c)$
$\Rightarrow 5\tan \frac{t}{2}=4\tan (4x+c)-3$
$\Rightarrow 5\tan (5x+3y)=4\tan (4x+c)-3$
Where $c=k$