What is the solution for 2y^{''}-7y^'+5y=-29 ?

The solution for 2y^{''}-7y^'+5y=-29 is y=c_{1}e^{(5t)/2}+c_{2}e^t-29/5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2y^{''}-7y^'+5y=-29

2 y^{^{''}} - 7 y^{^{'}} + 5 y = - 29

y = c_{1} e^{\frac{5 t}{2}} + c_{2} e^{t} - \frac{29}{5}

Solution steps

2 y^{''} (t) - 7 y^{'} (t) + 5 y = - 29

Solve linear ODE:

y = c_{1} e^{\frac{5 t}{2}} + c_{2} e^{t} - \frac{29}{5}

y = c_{1} e^{\frac{5 t}{2}} + c_{2} e^{t} - \frac{29}{5}

y^{^{''}} + y = t^{2} + 2, y (0) = 1, y^{^{'}} (0) = - 1

y^{^{''}} + 5 y^{^{'}} + 4 y = 3 x^{2}

y^{^{''}} + 4 y^{^{'}} + 3 y = 7 e^{- x}

y^{^{''}} + 8 y^{^{'}} + 16 y = e^{- 4 t}

y^{^{''}} + 9 y = sin (3 x) + e^{- x}

What is the solution for 2y^{''}-7y^'+5y=-29 ?
The solution for 2y^{''}-7y^'+5y=-29 is y=c_{1}e^{(5t)/2}+c_{2}e^t-29/5