What is the solution for x^{''}-3x^'+2x=t ?

The solution for x^{''}-3x^'+2x=t is x=c_{1}e^{2t}+c_{2}e^t+t/2+3/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^{''}-3x^'+2x=t

x^{^{''}} - 3 x^{^{'}} + 2 x = t

x = c_{1} e^{2 t} + c_{2} e^{t} + \frac{t}{2} + \frac{3}{4}

Solution steps

x^{''} (t) - 3 x^{'} (t) + 2 x = t

Solve linear ODE:

x = c_{1} e^{2 t} + c_{2} e^{t} + \frac{t}{2} + \frac{3}{4}

x = c_{1} e^{2 t} + c_{2} e^{t} + \frac{t}{2} + \frac{3}{4}

y^{^{'''}} + 8 y^{^{''}} + 20 y^{^{'}} = x^{2} - 4

y^{^{''}} - 4 y^{^{'}} - 21 = 0

y^{^{''}} + 256 y = sec (16 x)

y^{^{''}} + 7 y^{^{'}} + 6 y = sin (2 t), y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} - 6 y^{^{'}} - 5 y = 22 + 13 x - 5 x^{2}

What is the solution for x^{''}-3x^'+2x=t ?
The solution for x^{''}-3x^'+2x=t is x=c_{1}e^{2t}+c_{2}e^t+t/2+3/4