Find the general solution of the given differential equation.
y' + 3x2y = x2
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]"
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