What is the solution for y^{''}-(2α-12)y^'+(α^2-12α+20)y=0 ?

The solution for y^{''}-(2α-12)y^'+(α^2-12α+20)y=0 is y=c_{1}e^{(α-2)t}+c_{2}e^{(α-10)t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-(2α-12)y^'+(α^2-12α+20)y=0

y^{^{''}} - (2 α - 12) y^{^{'}} + (α^{2} - 12 α + 20) y = 0

y = c_{1} e^{(α - 2) t} + c_{2} e^{(α - 10) t}

Solution steps

y^{''} (t) - (2 α - 12) y^{'} (t) + (α^{2} - 12 α + 20) y = 0

Solve linear ODE:

y = c_{1} e^{(α - 2) t} + c_{2} e^{(α - 10) t}

y = c_{1} e^{(α - 2) t} + c_{2} e^{(α - 10) t}

\frac{d ^{2} x}{d t ^{2}} - a x = 0

y^{^{''}} - 3 y^{^{'}} - 4 y = 0, y (0) = a, y^{^{'}} (0) = - 10

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

y^{^{''}} - 4 y^{^{'}} + 3 y = 0, y (0) = - 1

\frac{d ^{2} x}{d t ^{2}} + 0.75 \frac{d x}{d t} + 0.3 x = 0

What is the solution for y^{''}-(2α-12)y^'+(α^2-12α+20)y=0 ?
The solution for y^{''}-(2α-12)y^'+(α^2-12α+20)y=0 is y=c_{1}e^{(α-2)t}+c_{2}e^{(α-10)t}