`y' + 3x^2y = x^2y^3` Solve the Bernoulli differential equation.

Expert Answers

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Multiply the above equation by `y^(-3)`



Taking the transformation  `v=y^(-2)`




Now the Bernoulli equation is transformed as ,



Now the above is a linear equation in the dependent variable v and independent variable y.

The integrating factor is n(x)=`e^(int(-6x^2dx))`










Let `t=x^3`





Substitute back `t=x^3`


Substitute back `v=y^(-2)`




 `y^2 = 1 / (1/3+Ce^(2x^3))`

`y = +-sqrt(3)/(Ce^(2x^3) + 1)`

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