What is the solution for y^'=3x^2y-8e^{(x^3-4x)} ?

The solution for y^'=3x^2y-8e^{(x^3-4x)} is y=2e^{-4x+x^3}+c_{1}e^{x^3}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=3x^2y-8e^{(x^3-4x)}

y^{^{'}} = 3 x^{2} y - 8 e^{(x^{3} - 4 x)}

y = 2 e^{- 4 x + x^{3}} + c_{1} e^{x^{3}}

Solution steps

y^{'} (x) = 3 x^{2} y - 8 e^{(x^{3} - 4 x)}

Solve linear ODE:

y = 2 e^{- 4 x + x^{3}} + c_{1} e^{x^{3}}

y = 2 e^{- 4 x + x^{3}} + c_{1} e^{x^{3}}

y^{^{''}} + 4 π^{2} y = cos^{3} (2 π x), y (0) = 1

f (x) = 5 - 5 x^{2}, (5, - 120)

x \to 5 lim (x^{2} + 2 x - 4)

\int e^{13 x} d x

\frac{1}{2 s + 6}

What is the solution for y^'=3x^2y-8e^{(x^3-4x)} ?
The solution for y^'=3x^2y-8e^{(x^3-4x)} is y=2e^{-4x+x^3}+c_{1}e^{x^3}