What is the solution for y^{''}-11y^'+24y=2,y(0)=0,y^'(0)=1 ?

The solution for y^{''}-11y^'+24y=2,y(0)=0,y^'(0)=1 is y=(1/3-1/12)e^{8t}-1/3 e^{3t}+1/12

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-11y^'+24y=2,y(0)=0,y^'(0)=1

y^{^{''}} - 11 y^{^{'}} + 24 y = 2, y (0) = 0, y^{^{'}} (0) = 1

y = (\frac{1}{3} - \frac{1}{12}) e^{8 t} - \frac{1}{3} e^{3 t} + \frac{1}{12}

Solution steps

y^{''} (t) - 11 y^{'} (t) + 24 y = 2

Solve linear ODE:

y = (\frac{1}{3} - \frac{1}{12}) e^{8 t} - \frac{1}{3} e^{3 t} + \frac{1}{12}

y = (\frac{1}{3} - \frac{1}{12}) e^{8 t} - \frac{1}{3} e^{3 t} + \frac{1}{12}

y^{^{''}} + 2 y^{^{'}} + 10 y = 10 x + 2

y^{^{'''}} + 3 y^{^{''}} - 7 y^{^{'}} + 2 y = 3 x + 10 sin (x)

y^{^{''}} + 2 y^{^{'}} + y = (2 + 1) t e^{(- t)}

y^{^{''}} - 4 y^{^{'}} + 4 y = x^{2} e^{2} x - e^{2} x

y^{^{'''}} - 2 y^{^{''}} + y^{^{'}} = 5 + 4 x e^{x}

What is the solution for y^{''}-11y^'+24y=2,y(0)=0,y^'(0)=1 ?
The solution for y^{''}-11y^'+24y=2,y(0)=0,y^'(0)=1 is y=(1/3-1/12)e^{8t}-1/3 e^{3t}+1/12