What is the solution for y^{''}-3y^{''}+3y^'-y=0 ?

The solution for y^{''}-3y^{''}+3y^'-y=0 is y=c_{1}e^{t/2}+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^{''}-3y^{''}+3y^'-y=0

y^{^{''}} - 3 y^{^{''}} + 3 y^{^{'}} - y = 0

y = c_{1} e^{\frac{t}{2}} + c_{2} e^{t}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{\frac{t}{2}} + c_{2} e^{t}

y = c_{1} e^{\frac{t}{2}} + c_{2} e^{t}

y^{^{''}} + 9 y = 0, y (0) = 2, y^{^{'}} (0) = 4

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

u^{^{''}} + 144 H (t) = 0, H (0) = 3, u^{^{'}} (0) = - 48

y^{^{'''}} - 7 y^{^{''}} + 19 y^{^{'}} - 13 y = 0

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

What is the solution for y^{''}-3y^{''}+3y^'-y=0 ?
The solution for y^{''}-3y^{''}+3y^'-y=0 is y=c_{1}e^{t/2}+c_{2}e^t