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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(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}

\frac{d}{d x} ((5 - e^{x})^{4})

\int_{0}^{\frac{1}{2}} \frac{x ^{2}}{1 - x ^{2}} d x

\frac{d y}{d x} = 3 y - y^{2}

\int \frac{ln ( 7 x )}{x} d x

y = cos (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