Question

Find the general solution of the differential equation tan y*(dy)/(dx)=sin(x+y) - sin(x-y).​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\rm{tan(y)\,\dfrac{dy}{dx}=sin(x+y)-sin(x-y)}$

Applying sin(C) - sin(D) formula, we get,

$\rm{\implies\,tan(y)\,\dfrac{dy}{dx}=2\,cos\left(\dfrac{x+y+x-y}{2}\right)\,sin\left(\dfrac{x+y-x+y}{2}\right)}$

$\rm{\implies\,tan(y)\,\dfrac{dy}{dx}=2\,cos\left(\dfrac{x+x}{2}\right)\,sin\left(\dfrac{y+y}{2}\right)}$

$\rm{\implies\,tan(y)\,\dfrac{dy}{dx}=2\,cos\left(\dfrac{2x}{2}\right)\,sin\left(\dfrac{2y}{2}\right)}$

$\rm{\implies\,tan(y)\,\dfrac{dy}{dx}=2\,cos(x)\,sin(y)}$

$\rm{\implies\,\dfrac{sin(y)}{cos(y)}\,\dfrac{dy}{dx}=2\,cos(x)\,sin(y)}$

$\rm{\implies\,\dfrac{dy}{cos(y)}=2\,cos(x)\,dx}$

Integrating both sides,

$\displaystyle\rm{\implies\,\int\dfrac{dy}{cos(y)}=2\int\,cos(x)\,dx}$

$\displaystyle\rm{\implies\,\int\,sec(y)\,dy=2\int\,cos(x)\,dx}$

$\displaystyle\rm{\implies\,\ln\left|sec(y)+tan(y)\right|=2\,sin(x)+C}$