Question

solve (xy sin xy +cos xy)ydx+(xy sin xy-cos xy)x dy=0​

Answer 1

SOLUTION

TO SOLVE

The differential equation

$\displaystyle\sf{(xy \sin xy + \cos xy)y dx + (xy \sin xy - \cos xy)x dy = 0}$

EVALUATION

Here the given differential equation is

$\displaystyle\sf{(xy \sin xy + \cos xy)y dx + (xy \sin xy - \cos xy)x dy = 0}$

$\displaystyle\sf{ \implies \: \frac{dy}{dx} = - \frac{y(xy \sin xy + \cos xy)}{x (xy \sin xy - \cos xy)} }$

$\sf{Let \: \: z = xy}$

Now Differentiating both sides with respect to x we get

$\displaystyle\sf{ \frac{dz}{dx} = y + x \frac{dy}{dx} }$

$\displaystyle\sf{ \implies \: \frac{dz}{dx} = y - \frac{xy(xy \sin xy + \cos xy)}{x (xy \sin xy - \cos xy)} }$

$\displaystyle\sf{ \implies \: \frac{dz}{dx} = y - \frac{y(z \sin z + \cos z)}{ (z \sin z - \cos z)} }$

$\displaystyle\sf{ \implies \: \frac{dz}{dx} = y \bigg[1- \frac{(z \sin z + \cos z)}{ (z \sin z - \cos z)} \bigg] }$

$\displaystyle\sf{ \implies \: \frac{dz}{dx} = y \bigg[ \frac{(z \sin z - \cos z - z \sin z - \cos z)}{ (z \sin z - \cos z)} \bigg] }$

$\displaystyle\sf{ \implies \: \frac{dz}{dx} = \frac{z}{x} \bigg[ \frac{ -2 \cos z }{ (z \sin z - \cos z)} \bigg] }$

$\displaystyle\sf{ \implies \: - \frac{(z \sin z - \cos z)}{ z\cos z } \: dz = \frac{2}{x}dx }$

$\displaystyle\sf{ \implies \: log |z \cos z| = 2 logx + logc}$

$\displaystyle\sf{ \implies \: log |z \cos z| = log c{x}^{2} }$

$\displaystyle\sf{ \implies \: |z \cos z| = c{x}^{2} }$

$\displaystyle\sf{ \implies \: | \: xy \cos xy \: | = c{x}^{2} }$

Where C is constant

FINAL ANSWER

Hence the required solution is

$\displaystyle\sf{ | \: xy \cos xy \: | = c{x}^{2} }$

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