Question

Find the general solution of the given differential equation.

y' + 3x²y = x²

Answer 1

Solution

From given differential equation

y'  = x²(1 – 3y)  => dy/(1 – 3y) = x²dx  => ∫dy/(1 – 3y) = ∫x²dx  =>  -ln|1 – 3y|/3 = x³/3 + C  =>  

$1-3y=De^{-x^3}$   =>  $y=\frac{1}{3}\left[1-De^{-x^3}\right]$

C and D are arbitrary constants.  

Answer

$y=\frac{1}{3}\left[1-De^{-x^3}\right]$