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Question

If dydx+xsin2y=x3cos2y,y(0)=0,then y(1)equal to

A
tan1(e)
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B
tan1(1e)
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C
tan1(e2)
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D
tan1(12e)
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Solution

The correct option is C tan1(e)
dydx+xsin2y=x3cos2y
divide both sides by cos2y
sec2ydydx+2sinycosycos2yx=x3
sec2ydydx+2tany.x=x3
set tany=tsec2ydyd2=d2dx
d2dx+2x.2=x3
linear differential equation
I.F=e2xdx=ex2
solution is t.ex2=ex2.x3dx+c
Let x2=n
xdx=12dn
y.ex2=C+12xendn
=C+12[nenen]=C+en2(n1)
zex2=C+12ex2(x21)
tany=Cex2+12(x21)
tany=Cex2+12(x21)

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