What is the solution for y^'=(-15x^2+2)(4y+7) ?

The solution for y^'=(-15x^2+2)(4y+7) is y=(e^{-20x^3+8x+c_{1}})/4-7/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^'=(-15x^2+2)(4y+7)

y^{^{'}} = (- 15 x^{2} + 2) (4 y + 7)

y = \frac{e ^{- 20 x^{3} + 8 x + c_{1}}}{4} - \frac{7}{4}

Solution steps

y^{'} (x) = (- 15 x^{2} + 2) (4 y + 7)

Solve separable ODE:

y = \frac{e ^{- 20 x^{3} + 8 x + c_{1}}}{4} - \frac{7}{4}

y = \frac{e ^{- 20 x^{3} + 8 x + c_{1}}}{4} - \frac{7}{4}

What is the solution for y^'=(-15x^2+2)(4y+7) ?
The solution for y^'=(-15x^2+2)(4y+7) is y=(e^{-20x^3+8x+c_{1}})/4-7/4