What is the solution for y^{''}-2y^'+y=e^x+x^2,y(0)=1,y^'(0)=0 ?

The solution for y^{''}-2y^'+y=e^x+x^2,y(0)=1,y^'(0)=0 is y=-5e^x+e^xx+(e^xx^2)/2+x^2+4x+6

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-2y^'+y=e^x+x^2,y(0)=1,y^'(0)=0

y^{^{''}} - 2 y^{^{'}} + y = e^{x} + x^{2}, y (0) = 1, y^{^{'}} (0) = 0

y = - 5 e^{x} + e^{x} x + \frac{e ^{x} x ^{2}}{2} + x^{2} + 4 x + 6

Solution steps

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

Solve linear ODE:

y = - 5 e^{x} + e^{x} x + \frac{e ^{x} x ^{2}}{2} + x^{2} + 4 x + 6

y = - 5 e^{x} + e^{x} x + \frac{e ^{x} x ^{2}}{2} + x^{2} + 4 x + 6

y^{^{''}} - 2 y^{^{'}} - 3 y = 2 cos (3 x)

\frac{d ^{4} y}{d x ^{4}} + 4 y = cos (2 x)

y^{^{''}} - 10 y^{^{'}} + 24 y = 90 e^{t}

y^{^{''}} + 9 y^{^{'}} = 8 sin (t), y (0) = 1, y^{^{'}} (0) = 1

y^{^{''}} + 2 y^{^{'}} + 17 = 0, y (0) = 1, y^{^{'}} (0) = - 1

What is the solution for y^{''}-2y^'+y=e^x+x^2,y(0)=1,y^'(0)=0 ?
The solution for y^{''}-2y^'+y=e^x+x^2,y(0)=1,y^'(0)=0 is y=-5e^x+e^xx+(e^xx^2)/2+x^2+4x+6