What is the solution for (dy)/(dx)= y/((x+1))+4x^2+4x ?

The solution for (dy)/(dx)= y/((x+1))+4x^2+4x is y=2x^2(x+1)-2(x+1)+c_{1}(x+1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(dy)/(dx)= y/((x+1))+4x^2+4x

\frac{d y}{d x} = \frac{y}{( x + 1 )} + 4 x^{2} + 4 x

y = 2 x^{2} (x + 1) - 2 (x + 1) + c_{1} (x + 1)

Solution steps

\frac{d y}{d x} = \frac{y}{( x + 1 )} + 4 x^{2} + 4 x

Solve linear ODE:

y = 2 x^{2} (x + 1) - 2 (x + 1) + c_{1} (x + 1)

y = 2 x^{2} (x + 1) - 2 (x + 1) + c_{1} (x + 1)

\frac{\partial}{\partial v} (cos (u))

\frac{\partial}{\partial y} (2 x y - 3 x^{2})

x \to 1 lim (\frac{2 x + 1}{x + 2})

\frac{d}{d x} ((9 x^{2} - 3 x - 10) y^{^{''}} + (x^{2} + x) y^{^{'}} + (2 x - 1) y)

\frac{\partial}{\partial x} (tan (6 + 2 x^{2} y^{3} z^{2}))

What is the solution for (dy)/(dx)= y/((x+1))+4x^2+4x ?
The solution for (dy)/(dx)= y/((x+1))+4x^2+4x is y=2x^2(x+1)-2(x+1)+c_{1}(x+1)