What is the solution for y^{''}-4y^'+4y=2x^2-4x+2 ?

The solution for y^{''}-4y^'+4y=2x^2-4x+2 is y=c_{1}e^{2x}+c_{2}xe^{2x}+(x^2)/2+1/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-4y^'+4y=2x^2-4x+2

y^{^{''}} - 4 y^{^{'}} + 4 y = 2 x^{2} - 4 x + 2

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + \frac{x ^{2}}{2} + \frac{1}{4}

Solution steps

y^{''} (x) - 4 y^{'} (x) + 4 y = 2 x^{2} - 4 x + 2

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + \frac{x ^{2}}{2} + \frac{1}{4}

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + \frac{x ^{2}}{2} + \frac{1}{4}

\frac{d ^{2} y}{d x ^{2}} - 6 \frac{d y}{d x} = 37 sin (x)

y^{^{''}} - 7 y^{^{'}} + 10 y = e^{x}

y^{^{''}} + k y = g - 6 t, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 8 y^{^{'}} + 16 y = 8 sin (4 x)

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

What is the solution for y^{''}-4y^'+4y=2x^2-4x+2 ?
The solution for y^{''}-4y^'+4y=2x^2-4x+2 is y=c_{1}e^{2x}+c_{2}xe^{2x}+(x^2)/2+1/4