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Question

I.F of dydx+xsin2y=x3cos2y is:

A
tany
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B
etany
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C
es˙my
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D
ex2
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Solution

The correct option is D ex2
Given the differential equation as dydx+xsin2y=x3cos2y
On dividing cos2y on both sides
sec2ydydx+x×sec2y×(2siny cosy)=x3
sec2ydydx+2x tany=x3
Let tany=P
On differentiating once,
sec2ydydx=dPdx............(substitute the above values to given D.E)
dPdx+2xP=x3
This is in the form of linear differential equation as dydx+p(X)y=q(X).
Then the solution of the equation is y×u(x)=((u(x)×q(x)dx)+C
where u(x)=e(p(x)dx) which is the integration factor of the equation.
On comparing the given equation with general equation.P(x)=2x,q(x)=x3
Integratingfactor=e(2xdx)=ex2

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