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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}-y^'=x^2

y^{^{''}} - y^{^{'}} = x^{2}

y = c_{1} e^{x} + c_{2} - \frac{x ^{3}}{3} - x^{2} - 2 x

Solution steps

y^{''} (x) - y^{'} (x) = x^{2}

Solve linear ODE:

y = c_{1} e^{x} + c_{2} - \frac{x ^{3}}{3} - x^{2} - 2 x

y = c_{1} e^{x} + c_{2} - \frac{x ^{3}}{3} - x^{2} - 2 x

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