What is the solution for (ye^x+xye^x)dx=-(xe^x+2)dy ?

The solution for (ye^x+xye^x)dx=-(xe^x+2)dy is y=(c_{1})/(e^xx+2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(ye^x+xye^x)dx=-(xe^x+2)dy

(y e^{x} + x y e^{x}) d x = - (x e^{x} + 2) d y

y = \frac{c _{1}}{e ^{x} x + 2}

Solution steps

(y e^{x} + x y e^{x}) d x = - (x e^{x} + 2) d y

Solve separable ODE:

y = \frac{c _{1}}{e ^{x} x + 2}

y = \frac{c _{1}}{e ^{x} x + 2}

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

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

y^{^{'}} = - 0.004 y + 50, y (0) = 0

y^{^{'}} = (x + 6) (y - 3), y (0) = 6

\frac{d x}{d y} = x^{6} - x

What is the solution for (ye^x+xye^x)dx=-(xe^x+2)dy ?
The solution for (ye^x+xye^x)dx=-(xe^x+2)dy is y=(c_{1})/(e^xx+2)