What is the solution for y^'-3/(x^2)y= 1/(x^2) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'-3/(x^2)y= 1/(x^2)

y^{^{'}} - \frac{3}{x ^{2}} y = \frac{1}{x ^{2}}

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

Solution steps

y^{'} (x) - \frac{3}{x ^{2}} y = \frac{1}{x ^{2}}

Solve separable ODE:

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

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

y^{^{'}} = \frac{t}{t ^{2} y + y}

n = 1 \sum \infty n e^{- \frac{n}{8}}

y^{^{''}} + 20 y^{^{'}} = 0, y (0) = - 5, y^{^{'}} (0) = 20

\frac{d ^{2}}{d x ^{2}} (x^{\frac{1}{11}})

\frac{\partial}{\partial x} (x e^{x^{2} + y^{2}})

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