What is the solution for y^{''}-5y^'+4y=0,y(0)=6,y^'(0)=2 ?

The solution for y^{''}-5y^'+4y=0,y(0)=6,y^'(0)=2 is y=(6-22/3)e^{4t}+22/3 e^t

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-5y^'+4y=0,y(0)=6,y^'(0)=2

y^{^{''}} - 5 y^{^{'}} + 4 y = 0, y (0) = 6, y^{^{'}} (0) = 2

y = (6 - \frac{22}{3}) e^{4 t} + \frac{22}{3} e^{t}

Solution steps

y^{''} (t) - 5 y^{'} (t) + 4 y = 0

Solve linear ODE:

y = (6 - \frac{22}{3}) e^{4 t} + \frac{22}{3} e^{t}

y = (6 - \frac{22}{3}) e^{4 t} + \frac{22}{3} e^{t}

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} = 0

y^{^{''}} + 0.8 y^{^{'}} + 100 y = 0, y (0) = 0, y^{^{'}} (0) = 4

18 y^{^{''}} + 72 y = 0

10 x^{^{''}} + \frac{1}{90} x = 0, x (0) = 100, x^{^{'}} (0) = 25

\frac{d ^{2} y}{d x ^{2}} + 6 \frac{d y}{d x} + 8 y = 0

What is the solution for y^{''}-5y^'+4y=0,y(0)=6,y^'(0)=2 ?
The solution for y^{''}-5y^'+4y=0,y(0)=6,y^'(0)=2 is y=(6-22/3)e^{4t}+22/3 e^t