What is the solution for y^'=(4x^4e^{y/x}+x^2y^2)/(x^3y) ?

The solution for y^'=(4x^4e^{y/x}+x^2y^2)/(x^3y) is y=-\W((4ln(x)+c_{1})/e)x-x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=(4x^4e^{y/x}+x^2y^2)/(x^3y)

y^{^{'}} = \frac{4 x ^{4} e ^{\frac{y}{x}} + x ^{2} y ^{2}}{x ^{3} y}

y = - W (\frac{4 ln ( x ) + c _{1}}{e}) x - x

Solution steps

y^{'} (x) = \frac{4 x ^{4} e ^{\frac{y}{x}} + x ^{2} y ^{2}}{x ^{3} y}

Solve using substitution:

y = - W (\frac{4 ln ( x ) + c _{1}}{e}) x - x

y = - W (\frac{4 ln ( x ) + c _{1}}{e}) x - x

\int x^{3} x^{2} - 81 d x

\int \frac{cot ^{3} ( x )}{5} d x

\frac{d}{d y} ((\frac{y}{8})^{4} + \frac{1}{4 y ^{2}})

x^{2} y^{^{''}} - 2 y = 0

x \to 1 lim (cos (π) x)

What is the solution for y^'=(4x^4e^{y/x}+x^2y^2)/(x^3y) ?
The solution for y^'=(4x^4e^{y/x}+x^2y^2)/(x^3y) is y=-\W((4ln(x)+c_{1})/e)x-x