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Question

Let  f:RR  be a differentiable function with f(0)=0.  If  y=f(x) satisfies the differential equation
dydx=(2+5y)(5y2)
then the value of lim  xf(x)  is


Solution

dydx=25y24125.dyy2(25)2=dx
Intregrating both side we get,
12512×25lny25y+25=x+c
Now, c=0 as f(0)=0.
Hence,  5y25y+2=e20xy=f(x)=25e20x1e20x+1
lim  xf(x)=lim  x25e20x1e20x+1=25=0.4

Mathematics

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