What is the solution for 2y^'=3xy+1 ?

The solution for 2y^'=3xy+1 is y=e^{(3x^2)/4}*integral of (e^{-(3x^2)/4})/(2)

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2y^'=3xy+1

2 y^{^{'}} = 3 x y + 1

y = e^{\frac{3 x ^{2}}{4}} \cdot \int \frac{e ^{- \frac{3 x ^{2}}{4}}}{2} d x

Solution steps

2 y^{'} (x) = 3 x y + 1

Solve linear ODE:

y = e^{\frac{3 x ^{2}}{4}} \cdot \int \frac{e ^{- \frac{3 x ^{2}}{4}}}{2} d x

y = e^{\frac{3 x ^{2}}{4}} \cdot \int \frac{e ^{- \frac{3 x ^{2}}{4}}}{2} d x

x \to 1 lim ((x^{2} - 2 x + 1) cos (e^{\frac{1}{x - 1}}))

x \to \infty lim ([2, x])

\int (sin^{2} (x) - cos^{2} (x)) d x

\frac{d ^{5}}{d x ^{5}} (\frac{1}{x})

\int e^{x} cot (e^{x}) d x

What is the solution for 2y^'=3xy+1 ?
The solution for 2y^'=3xy+1 is y=e^{(3x^2)/4}*integral of (e^{-(3x^2)/4})/(2)