What is the solution for y^{''}-6y^'+8y=2 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-6y^'+8y=2

y^{^{''}} - 6 y^{^{'}} + 8 y = 2

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

Solution steps

y^{''} (t) - 6 y^{'} (t) + 8 y = 2

Solve linear ODE:

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

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

y^{^{''}} + 3 y^{^{'}} - 4 = 0

y^{^{''}} + 4 y + 4 = x^{2} + 2 sin (x) cos (x)

y^{^{''}} - 2 y^{^{'}} + y = \frac{12 e ^{x}}{x ^{3}}

(D^{4} + D^{2}) y = x^{2} + 2 x + 4

\frac{d ^{2} y}{d x ^{2}} - \frac{d y}{d x} - 12 y = (x + 1) e^{x}

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