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Question

The solution of differential equation xdydx+y=x log x is

A
xy=x22log xx24
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B
xy=x22log xx24
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C
xy=x22log xx24+c,where is a constant
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D
None of the above
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Solution

The correct option is C xy=x22log xx24+c,where is a constant
The given equation is xdydx+y=x log x
Now you can see there is no multiplication of y and dydx, neither there is any transcendental function of y. Also dependent variable and y occur to its first degree. So this may be linear differential equation of first order.

If I write the given equation as dydx+yx=log x
Then it matches to dydx+P(x)y=Q(x)
Where P(x)=1x,Q(x)=log(x)
So this one is a linear differential equation of first order. Whose solution is.
y X (I.F) = Q(I.F.)dx + C
Where I.F. = ePdx
So I.F. = e1xdx=elog x=x
So solution is xy = x log x dx + C
xy=x22log xx24+C

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