Question

solve the differential equation

$\large\rm{ ( x + \tan \ y) dy = \sin \ 2y \ dx}$​

Answer 1

$\large\rm{ ( x + \tan \ y) dy = \sin \ 2y \ dx}$

$\large\rm{ \dfrac{dy}{dx} = \dfrac{x+ tan\:y}{\sin \:2y} }$

$\large\rm{ \dfrac{dy}{dx} = \dfrac{x}{\sin\:2y} + \dfrac{\tan\:y}{sin\:2y} }$

$\large\rm{ \dfrac{dy}{dx} = - \csc \: 2y \cdot x = \dfrac{1}{2} \sec^{2} y}$

$\large\rm$

since it is in form of dy/dx + Rx = S where R = - cosec 2y and S = ½ sec² y

$\large\rm{ \therefore I.F. = e^{\int Rdy} = e^{(- \csc \: 2y)dy} }$

$\large\rm{ = e^{ - \log | \csc \: 2y - \cot \: 2y|}}$

$\large\rm{ = e^{\log( \cot \ y)}}$

$\large\rm{ = \cot \ y}$

so solution for this differential equation is

$\large\rm{ x(I.F.) = \int S \cdot (I.F) dy + C}$

$\large\rm{ x( \cot \ y) = \int \dfrac{1}{2} \sec^{2} y \cdot ( \cot \ y)dy+C}$

$\large\bf { = \dfrac{1}{2} \ log | \cos \ 2y - \cot \ 2y| + C}$