What is the solution for 2y^'=(7e^x)/(2+y) ?

The solution for 2y^'=(7e^x)/(2+y) is y=-2+sqrt(c_{1)+7e^x+4},y=-sqrt(c_{1)+7e^x+4}-2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2y^'=(7e^x)/(2+y)

2 y^{^{'}} = \frac{7 e ^{x}}{2 + y}

y = - 2 + c_{1} + 7 e^{x} + 4, y = - c_{1} + 7 e^{x} + 4 - 2

Solution steps

2 y^{'} (x) = \frac{7 e ^{x}}{2 + y}

Solve separable ODE:

y = - 2 + c_{1} + 7 e^{x} + 4, y = - c_{1} + 7 e^{x} + 4 - 2

y = - 2 + c_{1} + 7 e^{x} + 4, y = - c_{1} + 7 e^{x} + 4 - 2

\frac{d x}{d y} = x y + x - 2 y - 2

\frac{d y}{d x} = \frac{( x y ^{2} - x )}{y}

y^{^{'}} = \frac{2 x}{e ^{2 y}}

x 1 + y^{2} d x - d y = 0

\frac{d y}{y} = \frac{d x}{( x - 9 ) ^{2} - 5 ^{2}}

What is the solution for 2y^'=(7e^x)/(2+y) ?
The solution for 2y^'=(7e^x)/(2+y) is y=-2+sqrt(c_{1)+7e^x+4},y=-sqrt(c_{1)+7e^x+4}-2