What is the solution for y^'=-(5y(2x+9)^2) ?

The solution for y^'=-(5y(2x+9)^2) is y=e^{-5((4x^3)/3+18x^2+81x)+c_{1}}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=-(5y(2x+9)^2)

y^{^{'}} = - (5 y (2 x + 9)^{2})

y = e^{- 5 (\frac{4 x ^{3}}{3} + 18 x^{2} + 81 x) + c_{1}}

Solution steps

y^{'} (x) = - (5 y (2 x + 9)^{2})

Solve separable ODE:

y = e^{- 5 (\frac{4 x ^{3}}{3} + 18 x^{2} + 81 x) + c_{1}}

y = e^{- 5 (\frac{4 x ^{3}}{3} + 18 x^{2} + 81 x) + c_{1}}

(\frac{d Q}{d t} + \frac{2}{300} Q) = 0.8

\frac{d y}{d x} = \frac{9 x ^{2}}{10 y ^{2}}

(x^{2} + 1) \frac{d y}{d x} + x y = x

cos (3 y) d y + sin (2 x) d x = 0

\frac{d x}{d t} + cos (t) x = cos (t), x (0) = 2

What is the solution for y^'=-(5y(2x+9)^2) ?
The solution for y^'=-(5y(2x+9)^2) is y=e^{-5((4x^3)/3+18x^2+81x)+c_{1}}