en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'+(1/x)y=8x^2

Solution

y^{^{'}} + (\frac{1}{x}) y = 8 x^{2}

Solution

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

Solution steps

y^{'} (x) + (\frac{1}{x}) y = 8 x^{2}

Solve linear ODE:

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

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

Popular Examples

integral from 2 to 4 of x^2y+y^2x+x+y

\int_{2}^{4} x^{2} y + y^{2} x + x + y d x

taylor 1/(1+x^{-1)}taylor

\frac{1}{1 + x ^{- 1}}

y^{''}-5y^'=0,y(0)=6,y^'(0)=-9

y^{^{''}} - 5 y^{^{'}} = 0, y (0) = 6, y^{^{'}} (0) = - 9

(\partial)/(\partial x)(ln(x^4-y^4))

\frac{\partial}{\partial x} (ln (x^{4} - y^{4}))

derivative of In^4x

\frac{d}{d x} (I n^{4} x)

Frequently Asked Questions (FAQ)

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