What is the solution for y^{''}-2y^'=8x+e^x ?

The solution for y^{''}-2y^'=8x+e^x is y=c_{1}e^{2x}+c_{2}-2x^2-2x-e^x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-2y^'=8x+e^x

y^{^{''}} - 2 y^{^{'}} = 8 x + e^{x}

y = c_{1} e^{2 x} + c_{2} - 2 x^{2} - 2 x - e^{x}

Solution steps

y^{''} (x) - 2 y^{'} (x) = 8 x + e^{x}

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} - 2 x^{2} - 2 x - e^{x}

y = c_{1} e^{2 x} + c_{2} - 2 x^{2} - 2 x - e^{x}

\int (x + 1)^{\frac{1}{3}} d x

\int \frac{8 x ^{3} + 2 x + 2}{2 x + 1} d x

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

2 x^{3} + x^{2} - x - 1, x^{3} + 2 x^{2} + 5 x - 1

(2 x - 2)^{4} (x^{2} + x + 1)^{5}

What is the solution for y^{''}-2y^'=8x+e^x ?
The solution for y^{''}-2y^'=8x+e^x is y=c_{1}e^{2x}+c_{2}-2x^2-2x-e^x