What is the solution for (D^3-4D^2+5D)y=0 ?

The solution for (D^3-4D^2+5D)y=0 is y=c_{1}+e^{2x}(c_{2}cos(x)+c_{3}sin(x))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(D^3-4D^2+5D)y=0

(D^{3} - 4 D^{2} + 5 D) y = 0

y = c_{1} + e^{2 x} (c_{2} cos (x) + c_{3} sin (x))

Solution steps

(D^{3} - 4 D^{2} + 5 D) y = 0

Solve linear ODE:

y = c_{1} + e^{2 x} (c_{2} cos (x) + c_{3} sin (x))

y = c_{1} + e^{2 x} (c_{2} cos (x) + c_{3} sin (x))

y^{^{''}} + 10 y^{^{'}} - 24 y = 0

4 y^{^{''''}} - 5 y^{^{''}} - 9 y = 0

y^{^{'''}} - 3 y^{^{''}} - 34 y^{^{'}} - 48 y = 0

y^{^{''}} - \frac{3}{2} y^{^{'}} - y = 0

y^{^{'''}} + 5 y^{^{''}} - 17 y^{^{'}} - 21 y = 0

What is the solution for (D^3-4D^2+5D)y=0 ?
The solution for (D^3-4D^2+5D)y=0 is y=c_{1}+e^{2x}(c_{2}cos(x)+c_{3}sin(x))