What is the solution for (dx)/(dy)=e^{2x+3y},x(0)=1 ?

The solution for (dx)/(dy)=e^{2x+3y},x(0)=1 is y=(ln(-3/2 e^{-2x}+c_{1}))/3

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(dx)/(dy)=e^{2x+3y},x(0)=1

\frac{d x}{d y} = e^{2 x + 3 y}, x (0) = 1

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

Solution steps

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

Solve separable ODE:

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

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

(3 x - 2) d x + (4 y + b) d y = 0

y^{^{'}} = \frac{( 3 t ^{2} )}{4 + 2 y}

y^{^{'}} = \frac{1}{2} (y^{2} - 1), y (0) = 6

y^{^{'}} = \frac{1}{2} (y^{2} - 1), y (0) = 4

y^{^{'}} = \frac{1}{2} (y^{2} - 1), y (0) = 2

What is the solution for (dx)/(dy)=e^{2x+3y},x(0)=1 ?
The solution for (dx)/(dy)=e^{2x+3y},x(0)=1 is y=(ln(-3/2 e^{-2x}+c_{1}))/3