What is the solution for y^{''}+4y^'-45y=0 ?

The solution for y^{''}+4y^'-45y=0 is y=c_{1}e^{5t}+c_{2}e^{-9t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}+4y^'-45y=0

y^{^{''}} + 4 y^{^{'}} - 45 y = 0

y = c_{1} e^{5 t} + c_{2} e^{- 9 t}

Solution steps

y^{''} (t) + 4 y^{'} (t) - 45 y = 0

Solve linear ODE:

y = c_{1} e^{5 t} + c_{2} e^{- 9 t}

y = c_{1} e^{5 t} + c_{2} e^{- 9 t}

\int \frac{1}{2 - e ^{3 x}} d x

\frac{d}{d x} ((x + 1) (x - 2)^{2})

n = 1 \sum \infty \frac{3}{n ^{2} + 5 n + 4}

x \to - 4 + lim (\frac{x + 5}{x + 4})

a re a 2 x^{3} - x^{2} - 5 x, - x^{2} + 3 x, 0, 2

What is the solution for y^{''}+4y^'-45y=0 ?
The solution for y^{''}+4y^'-45y=0 is y=c_{1}e^{5t}+c_{2}e^{-9t}