What is the solution for y^{''}-5y^'+4y=2e^x+x+1 ?

The solution for y^{''}-5y^'+4y=2e^x+x+1 is y=c_{1}e^{4x}+c_{2}e^x-(2e^xx)/3+x/4+9/16

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}-5y^'+4y=2e^x+x+1

y^{^{''}} - 5 y^{^{'}} + 4 y = 2 e^{x} + x + 1

y = c_{1} e^{4 x} + c_{2} e^{x} - \frac{2 e ^{x} x}{3} + \frac{x}{4} + \frac{9}{16}

Solution steps

y^{''} (x) - 5 y^{'} (x) + 4 y = 2 e^{x} + x + 1

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} e^{x} - \frac{2 e ^{x} x}{3} + \frac{x}{4} + \frac{9}{16}

y = c_{1} e^{4 x} + c_{2} e^{x} - \frac{2 e ^{x} x}{3} + \frac{x}{4} + \frac{9}{16}

What is the solution for y^{''}-5y^'+4y=2e^x+x+1 ?
The solution for y^{''}-5y^'+4y=2e^x+x+1 is y=c_{1}e^{4x}+c_{2}e^x-(2e^xx)/3+x/4+9/16