What is the solution for dy=(xy^2+3xy)dx ?

The solution for dy=(xy^2+3xy)dx is y=(3(e^{(-c_{1}-3x^2)/2}+1))/(e^{-c_{1)-3x^2}-1}

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dy=(xy^2+3xy)dx

d y = (x y^{2} + 3 x y) d x

y = \frac{3 ( e ^{\frac{- c _{1} - 3 x ^{2}}{2}} + 1 )}{e ^{- c_{1} - 3 x^{2}} - 1}

Solution steps

d y = (x y^{2} + 3 x y) d x

Solve Bernoulli ODE:

y = \frac{3 ( e ^{\frac{- c _{1} - 3 x ^{2}}{2}} + 1 )}{e ^{- c_{1} - 3 x^{2}} - 1}

y = \frac{3 ( e ^{\frac{- c _{1} - 3 x ^{2}}{2}} + 1 )}{e ^{- c_{1} - 3 x^{2}} - 1}

y^{^{''}} + 4 y^{^{'}} + 4 y = 25 cos (x)

n = 0 \sum \infty \frac{1}{3 ^{n} - n}

x^{y}

y = x^{2} 2 x - 7

\int (2 x^{2} + 1) e^{(x^{2})} d x

What is the solution for dy=(xy^2+3xy)dx ?
The solution for dy=(xy^2+3xy)dx is y=(3(e^{(-c_{1}-3x^2)/2}+1))/(e^{-c_{1)-3x^2}-1}