Let f:R→R be a differentiable function with f(0)=0. If y=f(x) satisfies the differential equation dydx=(2+5y)(5y−2)
then the value of limx→−∞f(x) is
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Solution
dydx=25y2−4⇒125.dyy2−(25)2=dx
Intregrating both side we get, 12512×25ln∣∣y−25y+25∣∣=x+c
Now, c=0 as f(0)=0.
Hence, 5y−25y+2=e20x⇒y=f(x)=25∣∣e20x−1e20x+1∣∣ limx→−∞f(x)=limx→−∞25∣∣e20x−1e20x+1∣∣=25=0.4