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Question

The solution of the differential equation xdydx+y=xcosx+sinx, given that y=1 when x=π2 is


A

y=sinx-cosx

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B

y=cosx

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C

y=sinx

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D

y=sinx+cosx

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Solution

The correct option is C

y=sinx


Explanation of the correct option.

Compute the required value.

Given :xdydx+y=xcosx+sinx

dydx+yx=cosx+sinx

Compare the equation with standard L.D.E.dydx+Py=Q,

P=1x and Q=cosx+sinx

Now,

I.F.=ePdx=e1xdx1xdx=lnx=elnx=x

Since the solution of L.D.E. is given by y(I.F.)=QI.F.dx+c

y.x=xcosx+sinxdx+c

y.x=x.sinx+c

Since y=1when x=π2

π2=π2+c

c=0

Therefore, y=sinx

Hence option C is the correct option.


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