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

The solution for (dy)/(dx)=e^{2x+3y},y(0)=0 is y=-(ln(-3/2 e^{2x}+5/2))/3

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

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

y = - \frac{ln ( - \frac{3}{2} e ^{2 x} + \frac{5}{2} )}{3}

Solution steps

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

Solve separable ODE:

y = - \frac{ln ( - \frac{3}{2} e ^{2 x} + \frac{5}{2} )}{3}

y = - \frac{ln ( - \frac{3}{2} e ^{2 x} + \frac{5}{2} )}{3}

\int_{6}^{\infty} x e^{- 2 x} d x

\int - 5 cos (x) d x

y^{^{''}} - 8 y^{^{'}} + 10 y = 0

\int 8 sec^{2} (x) d x

x \to 3 lim (f (x) (g (x)))

What is the solution for (dy)/(dx)=e^{2x+3y},y(0)=0 ?
The solution for (dy)/(dx)=e^{2x+3y},y(0)=0 is y=-(ln(-3/2 e^{2x}+5/2))/3