What is the solution for y^'=5x^4e^{-x^5}+e^x ?

The solution for y^'=5x^4e^{-x^5}+e^x is y=-e^{-x^5}+e^x+c_{1}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=5x^4e^{-x^5}+e^x

y^{^{'}} = 5 x^{4} e^{- x^{5}} + e^{x}

y = - e^{- x^{5}} + e^{x} + c_{1}

Solution steps

y^{'} (x) = 5 x^{4} e^{- x^{5}} + e^{x}

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

y = \int 5 x^{4} e^{- x^{5}} + e^{x} d x

\int 5 x^{4} e^{- x^{5}} + e^{x} d x = - e^{- x^{5}} + e^{x} + c_{1}

y = - e^{- x^{5}} + e^{x} + c_{1}

n = 0 \sum \infty n! 2^{n} n^{- n}

f (x) = 2 x^{3} + 3 x^{2} - 12 x + 8

y = \frac{x ^{x}}{sin ( x )}

x \to 2 lim (\frac{x ^{3} + 8}{x - 2})

\frac{\partial}{\partial y} (e^{x - 1} cos (y))

What is the solution for y^'=5x^4e^{-x^5}+e^x ?
The solution for y^'=5x^4e^{-x^5}+e^x is y=-e^{-x^5}+e^x+c_{1}