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rakibulalam109 rakibulalam109
  • 03-02-2017
  • Mathematics
Answered

(1+x^2)dy/dx+2xy-4x^2=0 bu using Bernoulli's Equation

Verified Answer :

LammettHash
LammettHash LammettHash
  • 03-02-2017

Not of Bernoulli type, but still linear.

\((1+x^2)\dfrac{\mathrm dy}{\mathrm dx}+2xy=4x^2\)

There's no need to find an integrating factor, since the left hand side already represents a derivative:

\(\dfrac{\mathrm d}{\mathrm dx}[(1+x^2)y]=(1+x^2)\dfrac{\mathrm dy}{\mathrm dx}+2xy\)

So, you have

\(\dfrac{\mathrm d}{\mathrm dx}[(1+x^2)y]=4x^2\)

and integrating both sides with respect to \(x\) yields

\((1+x^2)y=\displaystyle\int4x^2\,\mathrm dx\)
\((1+x^2)y=\dfrac43x^3+C\)
\(y=\dfrac{4x^3}{3(1+x^2)}+\dfrac C{1+x^2}\)

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