What is the solution for y^{''}-7y^'+6y=(x-2)e^x ?

The solution for y^{''}-7y^'+6y=(x-2)e^x is y=c_{1}e^{6x}+c_{2}e^x-(e^xx^2)/(10)+(9e^xx)/(25)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-7y^'+6y=(x-2)e^x

y^{^{''}} - 7 y^{^{'}} + 6 y = (x - 2) e^{x}

y = c_{1} e^{6 x} + c_{2} e^{x} - \frac{e ^{x} x ^{2}}{10} + \frac{9 e ^{x} x}{25}

Solution steps

y^{''} (x) - 7 y^{'} (x) + 6 y = (x - 2) e^{x}

Solve linear ODE:

y = c_{1} e^{6 x} + c_{2} e^{x} - \frac{e ^{x} x ^{2}}{10} + \frac{9 e ^{x} x}{25}

y = c_{1} e^{6 x} + c_{2} e^{x} - \frac{e ^{x} x ^{2}}{10} + \frac{9 e ^{x} x}{25}

y^{^{''}} - 2 y^{^{'}} + 37 y = e^{x} cos (6 x)

y^{^{''}} + 3 y^{^{'}} - 4 y = cos (2 x)

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

y^{^{''}} - y = 8

y^{^{''}} + y = H (t) - 2 H (t - \frac{b}{2}) + H (t - b)

What is the solution for y^{''}-7y^'+6y=(x-2)e^x ?
The solution for y^{''}-7y^'+6y=(x-2)e^x is y=c_{1}e^{6x}+c_{2}e^x-(e^xx^2)/(10)+(9e^xx)/(25)