Answer :

Given: differential equation xdy – (y + 2x^{2}) dx = 0

To find: the value of given differential equation

given: xdy – (y + 2x^{2}) dx = 0

Dividing throughout by dx,

Dividing throughout by x we get,

This is of the of the standard form, i.e.,

Where and Q=2x

And we know the integrating factor is

IF=e^{∫P dx}

Substituting the value of P in the IF, we get

But we know integration of is log x, substituting this in athe bove equation, we get

IF=e^{-log( x)}

Or,

But e^{log} ^{x}=x, so he above equation becomes,

IF=x^{-1}

So the general solution of a differential equation is

y(IF)=∫(Q×I.F)dx+C

Substituting the corresponding values in the above equation, we get

On integrating, we get

Or

y=2x^{2}+Cx

Hence this is the solution of the given differential equation

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