CUET Preparation Today
Solution of differential equation dy - sin x sin y dx = 0 is: |
$e^{cos\,x}.tan\frac{y}{2}=c$ $e^{cos\,x}.tany=c$ cos x . tan y = c cos x . sin y = c |
$e^{cos\,x}.tan\frac{y}{2}=c$ |
$\int\frac{dy}{\sin y}+\int\sin x dx⇒\log|\cot y-cosec y|= - \cos x + c$ $⇒\log|\tan \frac{y}{2}|=-\cos x + c ⇒ \tan \frac{y}{2}=e^{-\cos x}.e^c⇒e^{cos\,x}tan\frac{y}{2}=c$ |
