What is the solution for y^{''}-4y+5=0 ?

The solution for y^{''}-4y+5=0 is y=c_{1}e^{2t}+c_{2}e^{-2t}+5/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-4y+5=0

y^{^{''}} - 4 y + 5 = 0

y = c_{1} e^{2 t} + c_{2} e^{- 2 t} + \frac{5}{4}

Solution steps

y^{''} (t) - 4 y + 5 = 0

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{- 2 t} + \frac{5}{4}

y = c_{1} e^{2 t} + c_{2} e^{- 2 t} + \frac{5}{4}

4 y^{^{''}} - y = 1, y^{^{'}} (0) = \frac{1}{2}, y (0) = 0

y^{^{''}} + 6 y^{^{'}} + 9 y = \frac{- 10 e ^{- 3 t}}{t ^{2} + 1}

y^{^{''}} - y^{^{'}} - 20 = 0

\frac{d ^{2} y}{d x ^{2}} - 9 y = e^{3 x} cos (2 x) + e^{2 x}

y^{^{''}} + 9 y = cos (t)

What is the solution for y^{''}-4y+5=0 ?
The solution for y^{''}-4y+5=0 is y=c_{1}e^{2t}+c_{2}e^{-2t}+5/4