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Question

Solve the differential equation: dydxxsin2x=1xlogx

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Solution

dydxxsin2x=logxx
dydx.dx=logxx.dx+xsin2x.dx
Integrate on both sides, we get
dy=logxx.dx+xsin2x.dx
y=I1+I2 ........(1)
where I1=logxx.dx
Let t=logxdt=1xdx
So,I1=t.dt
=t22+c1
=(logx)22+c1 where t=logx
And, I2=xsin2x.dx
=12x(2sin2x).dx
=12x(1cos2x).dx
=12x.dx12xcos2x.dx
=12x2212xcos2x.dx
consider xcos2x.dx
u=xdu=dx
dv=cos2xdxv=sin2x2
xcos2x.dx=xsin2x2sin2x2dx
=xsin2x212sin2xdx
=xsin2x2+12cos2x2
I2=12x2212(xsin2x2+12cos2x2)
=x24xsin2x4cos2x8+c2
I=I1+I2
I=(logx)22+c1+x24xsin2x4cos2x8+c2
I=(logx)22+x24xsin2x4cos2x8+c where c=c1+c2

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