What is the solution for y^'+(6x+1/x)y=1 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^'+(6x+1/x)y=1

y^{^{'}} + (6 x + \frac{1}{x}) y = 1

y = \frac{1}{6 x} + \frac{c _{1}}{e ^{3 x^{2}} x}

Solution steps

y^{'} (x) + (6 x + \frac{1}{x}) y = 1

Solve linear ODE:

y = \frac{1}{6 x} + \frac{c _{1}}{e ^{3 x^{2}} x}

y = \frac{1}{6 x} + \frac{c _{1}}{e ^{3 x^{2}} x}

x \frac{d y}{d x} + 4 y = 5

\int 6 x e^{3 x^{2}} d x

a y^{^{''}} + b y^{^{'}} = 0

\int_{1}^{2} x - 1 d x

d er i v a t i v e f (x) = 6 6 x^{10} + 7 x^{6}

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