What is the solution for (dy)/(dx)=(3x^2+4x+2)/(2(y-1)),y(0)=-1 ?

The solution for (dy)/(dx)=(3x^2+4x+2)/(2(y-1)),y(0)=-1 is y=1-sqrt((x^2+2)(x+2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(dy)/(dx)=(3x^2+4x+2)/(2(y-1)),y(0)=-1

\frac{d y}{d x} = \frac{3 x ^{2} + 4 x + 2}{2 ( y - 1 )}, y (0) = - 1

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

Solution steps

\frac{d y}{d x} = \frac{3 x ^{2} + 4 x + 2}{2 ( y - 1 )}

Solve separable ODE:

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

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

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

\frac{d y}{d x} + 2 y = x

\int \frac{1}{144 - x ^{2}} d x

\int arccot (8 y) d y

\int \frac{1 - x}{1 + x} d x

What is the solution for (dy)/(dx)=(3x^2+4x+2)/(2(y-1)),y(0)=-1 ?
The solution for (dy)/(dx)=(3x^2+4x+2)/(2(y-1)),y(0)=-1 is y=1-sqrt((x^2+2)(x+2))