What is the solution for y^'-5y=te^{(-4t)} ?

The solution for y^'-5y=te^{(-4t)} is y=-(9t+1)/(81e^{4t)}+c_{1}e^{5t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = - \frac{9 t + 1}{81 e ^{4 t}} + c_{1} e^{5 t}

Solution steps

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

Solve linear ODE:

y = - \frac{9 t + 1}{81 e ^{4 t}} + c_{1} e^{5 t}

y = - \frac{9 t + 1}{81 e ^{4 t}} + c_{1} e^{5 t}

f (x) = \frac{1}{x ^{2}}, at x = 2

x y^{^{'}} = y + x^{2} sin (x), y (\frac{π}{3}) = 0

h (31 - \frac{1}{2} h)^{2}

\int_{0}^{1} \frac{sin ( x )}{x} d x

\int 4 ln (4 x^{2}) d x

What is the solution for y^'-5y=te^{(-4t)} ?
The solution for y^'-5y=te^{(-4t)} is y=-(9t+1)/(81e^{4t)}+c_{1}e^{5t}