The solution of x2dydx-xy-1-cos yx is

X^2dydx xy=1+cos yx dydx yx=1+cos yxx^2 Put y=vx⇒dydx=v+xdvdx ⇒ v+xdvdx v=1+cos vx^2 ⇒∫dv1+cos v=∫1x^3dx ⇒12∫dvcos^2 v2=∫1x^3dx ⇒12∫^2 v2· dv=∫1x^3dx ⇒12·tan v ...

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Byju's Answer

Standard XIII

Mathematics

Solving Homogeneous Differential Equations

The solution ...

Question

The solution of $x^{2} \frac{d y}{d x} - x y = 1 + cos \frac{y}{x}$ is

tan \frac{y}{2 x} = C - \frac{1}{2 x^{2}}

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tan \frac{y}{x} = C + \frac{1}{x}

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cos (\frac{y}{x}) = 1 + \frac{C}{x}

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x^{2} = (C + x^{2}) tan \frac{y}{x}

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Solution

The correct option is A $tan \frac{y}{2 x} = C - \frac{1}{2 x^{2}}$
$x^{2} \frac{d y}{d x} - x y = 1 + cos \frac{y}{x}$
$\frac{d y}{d x} - \frac{y}{x} = \frac{1 + cos \frac{y}{x}}{x^{2}}$
Put $y = v x \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x}$
$\Rightarrow v + x \frac{d v}{d x} - v = \frac{1 + cos v}{x^{2}}$
$\Rightarrow \int \frac{d v}{1 + cos v} = \int \frac{1}{x^{3}} d x$
$\Rightarrow \frac{1}{2} \int \frac{d v}{{cos}^{2} \frac{v}{2}} = \int \frac{1}{x^{3}} d x$
$\Rightarrow \frac{1}{2} \int {sec}^{2} \frac{v}{2} \cdot d v = \int \frac{1}{x^{3}} d x$
$\Rightarrow \frac{1}{2} \cdot tan \frac{v}{2} \cdot 2 = \frac{x^{- 3 + 1}}{- 3 + 1} + C$
$\Rightarrow tan \frac{y}{2 x} = C - \frac{1}{2 x^{2}}$

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The solution of x2dydx−xy=1+cosyx is

The correct option is A tany2x=C−12x2x2dydx−xy=1+cosyx dydx−yx=1+cosyxx2 Put y=vx⇒dydx=v+xdvdx ⇒v+xdvdx−v=1+cosvx2 ⇒∫dv1+cosv=∫1x3dx ⇒12∫dvcos2v2=∫1x3dx ⇒12∫sec2v2⋅dv=∫1x3dx ⇒12⋅tanv2⋅2=x−3+1−3+1+C ⇒tany2x=C−12x2

The solution of $x^{2} \frac{d y}{d x} - x y = 1 + cos \frac{y}{x}$ is