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Q.

Solution of xy-dydx=y3e-x2 is

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a

ex2=y2(2x-c)

b

y2=ex2(2x-c)

c

y2=e-x2(2x-c)

d

e-x2=y2(2x-c)

answer is B.

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Detailed Solution

-dydx+xy=y3 e-x2  -1y3 dydx+1y2x=e-x2  Put 1y2=t-2y-3 dydx=dtdx  -1y3 dydx=12 dtdx

Then we get,

12 dtdx+xt=e-x2  dtdx+2xt=2e-x2

 Linear Diff. eqn.

  dydx+y P(x)=Q(x)I.F=eP(x) dxy·IF=Q(x)·IFI·F=e2x dx =ex2tex2=2 e-x2 ex2 dx =2e0 dx  1y2 ex2=2x+cex2=2xy2+cy2 

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