What is the solution for x^{''}-14x^'+49x=4te^{7t} ?

The solution for x^{''}-14x^'+49x=4te^{7t} is x=c_{1}e^{7t}+c_{2}te^{7t}+(2e^{7t}t^3)/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^{''}-14x^'+49x=4te^{7t}

x^{^{''}} - 14 x^{^{'}} + 49 x = 4 t e^{7 t}

x = c_{1} e^{7 t} + c_{2} t e^{7 t} + \frac{2 e ^{7 t} t ^{3}}{3}

Solution steps

x^{''} (t) - 14 x^{'} (t) + 49 x = 4 t e^{7 t}

Solve linear ODE:

x = c_{1} e^{7 t} + c_{2} t e^{7 t} + \frac{2 e ^{7 t} t ^{3}}{3}

x = c_{1} e^{7 t} + c_{2} t e^{7 t} + \frac{2 e ^{7 t} t ^{3}}{3}

x^{3} + 4 x

\frac{\partial}{\partial y} (x^{2} - x + g (y))

\frac{d y}{d x} = (\frac{y}{( x + 2 y )})

\int ∣ 3 - x ∣ d x

\frac{d}{d x} (\frac{1}{x} - \frac{1}{x ^{2}})

What is the solution for x^{''}-14x^'+49x=4te^{7t} ?
The solution for x^{''}-14x^'+49x=4te^{7t} is x=c_{1}e^{7t}+c_{2}te^{7t}+(2e^{7t}t^3)/3