What is the solution for (x+2)^2y^'+3(x+2)y=4 ?

The solution for (x+2)^2y^'+3(x+2)y=4 is y=(2(x^2+4x))/(x^3+6x^2+12x+8)+(c_{1})/(x^3+6x^2+12x+8)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

(x+2)^2y^'+3(x+2)y=4

(x + 2)^{2} y^{^{'}} + 3 (x + 2) y = 4

y = \frac{2 ( x ^{2} + 4 x )}{x ^{3} + 6 x ^{2} + 12 x + 8} + \frac{c _{1}}{x ^{3} + 6 x ^{2} + 12 x + 8}

Solution steps

(x + 2)^{2} y^{'} (x) + 3 (x + 2) y = 4

Solve linear ODE:

y = \frac{2 ( x ^{2} + 4 x )}{x ^{3} + 6 x ^{2} + 12 x + 8} + \frac{c _{1}}{x ^{3} + 6 x ^{2} + 12 x + 8}

y = \frac{2 ( x ^{2} + 4 x )}{x ^{3} + 6 x ^{2} + 12 x + 8} + \frac{c _{1}}{x ^{3} + 6 x ^{2} + 12 x + 8}

\frac{d y}{d x} = 2 y + y^{5}, y (0) = 1

\frac{d}{d x} (ln (2 x^{2} - 3 x + 1))

t an g e n t f (x) = x x, at x = 9

x \to \infty lim (\frac{7.5 x + 120000}{x})

\int θ sin (θ) cos (θ) d θ

What is the solution for (x+2)^2y^'+3(x+2)y=4 ?
The solution for (x+2)^2y^'+3(x+2)y=4 is y=(2(x^2+4x))/(x^3+6x^2+12x+8)+(c_{1})/(x^3+6x^2+12x+8)