What is the solution for 2x^2y^'=y^'+5xe^{-y} ?

The solution for 2x^2y^'=y^'+5xe^{-y} is y=ln(5/4 ln(2x^2-1)+c_{1})

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = ln (\frac{5}{4} ln (2 x^{2} - 1) + c_{1})

Solution steps

2 x^{2} y^{'} (x) = y^{'} (x) + 5 x e^{- y}

Solve separable ODE:

y = ln (\frac{5}{4} ln (2 x^{2} - 1) + c_{1})

y = ln (\frac{5}{4} ln (2 x^{2} - 1) + c_{1})

\int \frac{4}{sec ( x ) e ^{s i n (x)}} d x

y = (1 + ln (x))^{s i n (x)}

\frac{d}{d x} ((5 x)^{\frac{1}{3}})

\int_{0}^{8} x (\frac{1}{8} x) d x

\frac{d}{d x} ((x^{3} - 5)^{5} (x^{5} + 7))

What is the solution for 2x^2y^'=y^'+5xe^{-y} ?
The solution for 2x^2y^'=y^'+5xe^{-y} is y=ln(5/4 ln(2x^2-1)+c_{1})