Question

cosx dy =y(sinx_y)dx​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\sf{cos(x)\,dy=y\left(sin(x)-y\right)dx}$

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

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

$\sf{\implies\,\dfrac{dy}{dx}=y\,tan(x)-y^2\,sec(x)}$

$\sf{\implies\,\dfrac{dy}{dx}-y\,tan(x)=-y^2\,sec(x)}$

$\sf{\implies\,\dfrac{1}{y^2}\,\dfrac{dy}{dx}-\dfrac{1}{y}\,tan(x)=-\,sec(x)}$

$\bf{\mapsto\,\,Let\,\,\,-\dfrac{1}{y}=v}$

$\bf{\mapsto\,\,\dfrac{1}{y^2}\,\dfrac{dy}{dx}=\dfrac{dv}{dx}}$

So,

$\sf{\implies\,\dfrac{dv}{dx}+v\,tan(x)=-\,sec(x)}$

This is a linear differential equation, so,

$\bf{I.F.=\displaystyle\,e^{\displaystyle\int\,tan(x)\,dx}}$

$\bf{\implies\,I.F.=\displaystyle\,e^{\displaystyle\ln|sec(x)|}}$

$\bf{\implies\,I.F.=sec(x)}$

So, our solution is

$\sf{\displaystyle\,sec(x)\cdot\,v=-\int\,sec(x)\cdot\,sec(x)\,dx}$

$\sf{\displaystyle\implies\,sec(x)\cdot\,v=-\int\,sec^2(x)\,dx}$

$\sf{\implies\,sec(x)\cdot\,v=-tan(x)+c}$

$\sf{\implies\,-\,sec(x)\cdot\,\dfrac{1}{y}=-tan(x)+c}$

$\sf{\implies\,sec(x)\cdot\,\dfrac{1}{y}=tan(x)-c}$

Putting - c = k

$\sf{\implies\,sec(x)\cdot\,\dfrac{1}{y}=tan(x)+k}$

$\sf{\implies\,\dfrac{1}{y}=\dfrac{tan(x)}{sec(x)}+\dfrac{k}{sec(x)}}$

$\sf{\implies\,\dfrac{1}{y}=sin(x)+k\,cos(x)}$

$\sf{\implies\,y=\dfrac{1}{sin(x)+k\,cos(x)}}$