What is the solution for y^{''}-2y^'+y=-5e^x+8x^2 ?

The solution for y^{''}-2y^'+y=-5e^x+8x^2 is y=c_{1}e^x+c_{2}xe^x-(5e^xx^2)/2+8x^2+32x+48

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}-2y^'+y=-5e^x+8x^2

y^{^{''}} - 2 y^{^{'}} + y = - 5 e^{x} + 8 x^{2}

y = c_{1} e^{x} + c_{2} x e^{x} - \frac{5 e ^{x} x ^{2}}{2} + 8 x^{2} + 32 x + 48

Solution steps

y^{''} (x) - 2 y^{'} (x) + y = - 5 e^{x} + 8 x^{2}

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} - \frac{5 e ^{x} x ^{2}}{2} + 8 x^{2} + 32 x + 48

y = c_{1} e^{x} + c_{2} x e^{x} - \frac{5 e ^{x} x ^{2}}{2} + 8 x^{2} + 32 x + 48

What is the solution for y^{''}-2y^'+y=-5e^x+8x^2 ?
The solution for y^{''}-2y^'+y=-5e^x+8x^2 is y=c_{1}e^x+c_{2}xe^x-(5e^xx^2)/2+8x^2+32x+48