What is the solution for y^{''}-8y^'+15=0 ?

The solution for y^{''}-8y^'+15=0 is y=c_{1}e^{8t}+c_{2}+(15t)/8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-8y^'+15=0

y^{^{''}} - 8 y^{^{'}} + 15 = 0

y = c_{1} e^{8 t} + c_{2} + \frac{15 t}{8}

Solution steps

y^{''} (t) - 8 y^{'} (t) + 15 = 0

Solve linear ODE:

y = c_{1} e^{8 t} + c_{2} + \frac{15 t}{8}

y = c_{1} e^{8 t} + c_{2} + \frac{15 t}{8}

f (x) = tan (2 x - x^{3})

\frac{d}{d x} (cos (x) - 3 tan (x))

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

\frac{d}{d x} (\frac{80}{x ^{2}})

\frac{\partial}{\partial y} (9 x^{8 y})

What is the solution for y^{''}-8y^'+15=0 ?
The solution for y^{''}-8y^'+15=0 is y=c_{1}e^{8t}+c_{2}+(15t)/8