What is the solution for y^{''}-8y^'+16y=0 ?

The solution for y^{''}-8y^'+16y=0 is y=c_{1}e^{4t}+c_{2}te^{4t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-8y^'+16y=0

y^{^{''}} - 8 y^{^{'}} + 16 y = 0

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

Solution steps

y^{''} (t) - 8 y^{'} (t) + 16 y = 0

Solve linear ODE:

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

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

\int \frac{x ^{2} + 1}{( x - 1 ) ^{3}} d x

\frac{d}{d x} (\frac{sin ( x )}{cos ( x ^{2} )})

\int x^{4} \cdot e^{- 6 x^{5}} d x

\frac{d}{d x} (tan (x^{7}))

\frac{d}{d x} (8 sin^{3} (x))

What is the solution for y^{''}-8y^'+16y=0 ?
The solution for y^{''}-8y^'+16y=0 is y=c_{1}e^{4t}+c_{2}te^{4t}