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Solution of diffrential equation (x \sin x/y + y )dy = (\sin x/y )\cdot ydx  is 

  • Option 1)

    ln(y) + \sin (x/y) =C

  • Option 2)

    ln|y |+ \cos (x/y) =C

  • Option 3)

    ln|y| - \sin (x/y) =C

  • Option 4)

    ln|y |- \cos (x/y) =C

 

Answers (1)

best_answer

As we have learned

General form of Variable Separation -

d\left ( \frac{x}{y} \right )= \frac{ydx-xdy}{y^{2}}

-

 

 Given equation becomes  --

\sin x/y (xdy -ydx) + ydy = 0

\Rightarrowydy= (ydx-xdy)\cdot \sin (x/y)

dividing both by y^2 we get 

dy/y = \frac{ydx-xdy}{y^2}\cdot \sin (x/y)

\Rightarrow dy/y = d(x/y)\cdot \sin x/y

\Rightarrow \int dy/y -\int d(x/y)\cdot \sin x/y =C

ln |y| + \cos x/y = C

 

 

 

 

 


Option 1)

ln(y) + \sin (x/y) =C

Option 2)

ln|y |+ \cos (x/y) =C

Option 3)

ln|y| - \sin (x/y) =C

Option 4)

ln|y |- \cos (x/y) =C

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