What is the solution for y^{''}-4y^'+3y=0,y(0)=-2,y(0)=-14 ?

The solution for y^{''}-4y^'+3y=0,y(0)=-2,y(0)=-14 is y=(-2-c_{2})e^{3t}+c_{2}e^t

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-4y^'+3y=0,y(0)=-2,y(0)=-14

y^{^{''}} - 4 y^{^{'}} + 3 y = 0, y (0) = - 2, y (0) = - 14

y = (- 2 - c_{2}) e^{3 t} + c_{2} e^{t}

Solution steps

y^{''} (t) - 4 y^{'} (t) + 3 y = 0

Solve linear ODE:

y = (- 2 - c_{2}) e^{3 t} + c_{2} e^{t}

y = (- 2 - c_{2}) e^{3 t} + c_{2} e^{t}

2 y^{^{''}} + 18 y = 0, y (0) = 2, y^{^{'}} (0) = 15

y^{^{'''''}} + 20 y^{^{'''}} = 0

\frac{5}{8} x^{^{''}} + 40 x = 0, x (0) = \frac{1}{3}, x^{^{'}} (0) = 0

y^{^{''}} - 2 y^{^{'}} - 8 y = 0, y (0) = a, y^{^{'}} (0) = 10

y^{^{''}} + 2 α y^{^{'}} + α^{2} y = 0

What is the solution for y^{''}-4y^'+3y=0,y(0)=-2,y(0)=-14 ?
The solution for y^{''}-4y^'+3y=0,y(0)=-2,y(0)=-14 is y=(-2-c_{2})e^{3t}+c_{2}e^t