What is the solution for (2x+ye^{y/x})dx-xe^{y/x}dy=0,y(1)=4 ?

The solution for (2x+ye^{y/x})dx-xe^{y/x}dy=0,y(1)=4 is y=ln(2ln(x)+c_{1})x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(2x+ye^{y/x})dx-xe^{y/x}dy=0,y(1)=4

(2 x + y e^{\frac{y}{x}}) d x - x e^{\frac{y}{x}} d y = 0, y (1) = 4

y = ln (2 ln (x) + c_{1}) x

Solution steps

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

Solve using substitution:

y = ln (2 ln (x) + c_{1}) x

y = ln (2 ln (x) + c_{1}) x

x \to \infty lim (\frac{1}{5 ^{\frac{1}{x}}})

\frac{1}{2} sin^{2} (x)

y = x^{2} - 5 x + 5

\frac{d}{d x} (x^{2} + \frac{c}{x ^{2}})

\frac{\partial}{\partial x} (x^{2} + y^{2} - 4)

What is the solution for (2x+ye^{y/x})dx-xe^{y/x}dy=0,y(1)=4 ?
The solution for (2x+ye^{y/x})dx-xe^{y/x}dy=0,y(1)=4 is y=ln(2ln(x)+c_{1})x