What is the solution for y^{''}-6y^'+8y=0 ?

The solution for y^{''}-6y^'+8y=0 is y=c_{1}e^{4t}+c_{2}e^{2t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-6y^'+8y=0

y^{^{''}} - 6 y^{^{'}} + 8 y = 0

y = c_{1} e^{4 t} + c_{2} e^{2 t}

Solution steps

y^{''} (t) - 6 y^{'} (t) + 8 y = 0

Solve linear ODE:

y = c_{1} e^{4 t} + c_{2} e^{2 t}

y = c_{1} e^{4 t} + c_{2} e^{2 t}

\int sin^{7} (x) cos^{2} (x) d x

y = lo g_{5} (x)

\frac{d}{d x} (lo g_{10} (2 x^{2} + 5 x + 2))

\frac{3 x - y}{x}

\frac{\partial}{\partial x} (x^{0.2} y^{0.4})

What is the solution for y^{''}-6y^'+8y=0 ?
The solution for y^{''}-6y^'+8y=0 is y=c_{1}e^{4t}+c_{2}e^{2t}