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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\int cos^{4} (3 t) d t

\int_{- π}^{π} x d x

\frac{d}{d x} (\frac{( x ^{2} )}{5 + 8 x})

\frac{d}{d x} (\frac{20}{x ^{6}})

\frac{d ^{2} y}{d x ^{2}} + 5 \frac{d y}{d x} + 6 y = 12 e^{x}

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