What is the solution for 2x^2y^{''}+5xy^'+y=18x+4 ?

The solution for 2x^2y^{''}+5xy^'+y=18x+4 is y=(c_{1}}{sqrt(x)}+\frac{c_{2})/x+3x+4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2x^2y^{''}+5xy^'+y=18x+4

2 x^{2} y^{^{''}} + 5 x y^{^{'}} + y = 18 x + 4

y = \frac{c _{1}}{x} + \frac{c _{2}}{x} + 3 x + 4

Solution steps

2 x^{2} y^{''} (x) + 5 x y^{'} (x) + y = 18 x + 4

Solve linear ODE:

y = \frac{c _{1}}{x} + \frac{c _{2}}{x} + 3 x + 4

y = \frac{c _{1}}{x} + \frac{c _{2}}{x} + 3 x + 4

y^{^{'}} + y^{2} sin (x) = 0

x \to \infty lim (sin (1))

\frac{d y}{d x} = \frac{2 y}{2 x + 1}

10 x^{2} + 2 x

\int 8 t^{2} d t

What is the solution for 2x^2y^{''}+5xy^'+y=18x+4 ?
The solution for 2x^2y^{''}+5xy^'+y=18x+4 is y=(c_{1}}{sqrt(x)}+\frac{c_{2})/x+3x+4