Question

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

Answer 1

Not of Bernoulli type, but still linear.


`(1+x^2)(\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:


`(\mathrm d)/(\mathrm dx)[(1+x^2)y]=(1+x^2)(\mathrm dy)/(\mathrm dx)+2xy`

So, you have


`(\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=\frac43x^3+C`

`y=(4x^3)/(3(1+x^2))+\frac C{1+x^2}`