What is the solution for 8y^{''}-9y^'+y=0,y(0)=8,y^'(0)= 1/8 ?

The solution for 8y^{''}-9y^'+y=0,y(0)=8,y^'(0)= 1/8 is y=-e^t+9e^{t/8}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

8y^{''}-9y^'+y=0,y(0)=8,y^'(0)= 1/8

8 y^{^{''}} - 9 y^{^{'}} + y = 0, y (0) = 8, y^{^{'}} (0) = \frac{1}{8}

y = - e^{t} + 9 e^{\frac{t}{8}}

Solution steps

8 y^{''} (t) - 9 y^{'} (t) + y = 0

Solve linear ODE:

y = - e^{t} + 9 e^{\frac{t}{8}}

y = - e^{t} + 9 e^{\frac{t}{8}}

y^{^{''''}} - 2 y^{^{'''}} - 10 y^{^{''}} + 6 y^{^{'}} + 45 y = 0

16 \frac{d ^{4} y}{d x ^{4}} + 16 \frac{d ^{2} y}{d x ^{2}} + 4 y = 0

y^{^{''}} + y^{^{'}} = 6 y

y^{^{''}} + 4 y = 0, y (\frac{π}{4}) = - 2, y (0) = 1

y^{^{''}} + 2 y^{^{'}} + y = 0, y (0) = - 1, y^{^{'}} (0) = - 2

What is the solution for 8y^{''}-9y^'+y=0,y(0)=8,y^'(0)= 1/8 ?
The solution for 8y^{''}-9y^'+y=0,y(0)=8,y^'(0)= 1/8 is y=-e^t+9e^{t/8}