What is the solution for y^{''}-3y^'+2y=x ?

The solution for y^{''}-3y^'+2y=x is y=c_{1}e^{2x}+c_{2}e^x+x/2+3/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-3y^'+2y=x

y^{^{''}} - 3 y^{^{'}} + 2 y = x

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

Solution steps

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

Solve linear ODE:

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

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

5 y^{^{''}} - 10 y^{^{'}} + 5 y = e^{x} ln (x)

y^{^{''}} + 5 y^{^{'}} + 6 y = 3 t^{2} + 4

y^{^{''}} + y = sec^{3} (x), y (0) = 1, y^{^{'}} (0) = \frac{1}{2}

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

y^{^{''}} + y = 2, y (0) = 2, y^{^{'}} (0) = 3

What is the solution for y^{''}-3y^'+2y=x ?
The solution for y^{''}-3y^'+2y=x is y=c_{1}e^{2x}+c_{2}e^x+x/2+3/4