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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

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

Solution steps

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

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}

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