What is the solution for y^{''}-2/5 y=-x^2 ?

The solution for y^{''}-2/5 y=-x^2 is y=c_{1}e^{sqrt(2/5)x}+c_{2}e^{-sqrt(2/5)x}+(5x^2)/2+25/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}-2/5 y=-x^2

y^{^{''}} - \frac{2}{5} y = - x^{2}

y = c_{1} e^{\frac{2}{5} x} + c_{2} e^{- \frac{2}{5} x} + \frac{5 x ^{2}}{2} + \frac{25}{2}

Solution steps

y^{''} (x) - \frac{2}{5} y = - x^{2}

Solve linear ODE:

y = c_{1} e^{\frac{2}{5} x} + c_{2} e^{- \frac{2}{5} x} + \frac{5 x ^{2}}{2} + \frac{25}{2}

y = c_{1} e^{\frac{2}{5} x} + c_{2} e^{- \frac{2}{5} x} + \frac{5 x ^{2}}{2} + \frac{25}{2}

What is the solution for y^{''}-2/5 y=-x^2 ?
The solution for y^{''}-2/5 y=-x^2 is y=c_{1}e^{sqrt(2/5)x}+c_{2}e^{-sqrt(2/5)x}+(5x^2)/2+25/2