What is the solution for x^{''}-8x^'+17x=0 ?

The solution for x^{''}-8x^'+17x=0 is x=e^{4t}(c_{1}cos(t)+c_{2}sin(t))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^{''}-8x^'+17x=0

x^{^{''}} - 8 x^{^{'}} + 17 x = 0

x = e^{4 t} (c_{1} cos (t) + c_{2} sin (t))

Solution steps

x^{''} (t) - 8 x^{'} (t) + 17 x = 0

Solve linear ODE:

x = e^{4 t} (c_{1} cos (t) + c_{2} sin (t))

x = e^{4 t} (c_{1} cos (t) + c_{2} sin (t))

\frac{d}{d x} (e^{\frac{x}{( x - 6 )}})

\frac{d}{d x} (ln (1 - \frac{3}{x}))

x \to 2 lim (\frac{x ^{2} - 2 x}{x})

\int \frac{2}{x ^{3}} d x

\int \frac{( 3 x ^{2} + 6 )}{( x ^{2} - 2 x + 2 ) ^{2}} d x

What is the solution for x^{''}-8x^'+17x=0 ?
The solution for x^{''}-8x^'+17x=0 is x=e^{4t}(c_{1}cos(t)+c_{2}sin(t))