What is the solution for dy=(e^{3x+2y})dx ?

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

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dy=(e^{3x+2y})dx

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

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

Solution steps

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

Solve separable ODE:

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

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

x \frac{d y}{d x} = e^{10 y + x}

\frac{d y}{d x} = (x y)

(x + 1) y^{^{'}} + y^{2} = 0

\frac{d x}{d t} + \frac{x}{100} = 75

(x^{2} - 3) \frac{d y}{d x} = x (y^{2} + 16), y (2) = 4

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