Question

dy/dx+y.tanx=y^2secx
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Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{\dfrac{dy}{dx}+y\,tan(x)=y^2\,sec(x)}$

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

$\sf{Put\,\,\,\dfrac{1}{y}=v}$

$\sf{\implies-\dfrac{1}{y^2}\dfrac{dy}{dx}=\dfrac{dv}{dx}}$

So,

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

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

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

$\sf{=e^{ln|cos(x)|}}$

$\sf{=cos(x)}$

Now,

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

$\sf{\implies\,v\cdot\,cos(x)=\displaystyle\int(-1)\,dx}$

$\sf{\implies\,v\cdot\,cos(x)=-x+C}$

$\sf{\implies\,\dfrac{cos(x)}{y}=-x+C}$