What is the solution for y^'=e^{3x-2y}+x^2e^{-2y} ?

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

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y^'=e^{3x-2y}+x^2e^{-2y}

y^{^{'}} = e^{3 x - 2 y} + x^{2} e^{- 2 y}

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

Solution steps

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

Solve separable ODE:

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

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

\frac{d t}{d x} + \frac{3}{x} = - \frac{1}{x ^{2}}

\frac{d y}{d x} = 0.06 - \frac{y}{1350}

y^{^{'}} - 5 y = 10, y (0) = 0

f^{^{'}} (p) - f (p) = 0

((e^{3 x} + 3) e^{y}) d y - ((e^{y} - 4) e^{3 x}) d x = 0

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