What is the solution for 4y^{''}-4y^'+y=24e^{t/2} ?

The solution for 4y^{''}-4y^'+y=24e^{t/2} is y=c_{1}e^{t/2}+c_{2}e^{t/2}t+3e^{t/2}t^2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4y^{''}-4y^'+y=24e^{t/2}

4 y^{^{''}} - 4 y^{^{'}} + y = 24 e^{\frac{t}{2}}

y = c_{1} e^{\frac{t}{2}} + c_{2} e^{\frac{t}{2}} t + 3 e^{\frac{t}{2}} t^{2}

Solution steps

4 y^{''} (t) - 4 y^{'} (t) + y = 24 e^{\frac{t}{2}}

Solve linear ODE:

y = c_{1} e^{\frac{t}{2}} + c_{2} e^{\frac{t}{2}} t + 3 e^{\frac{t}{2}} t^{2}

y = c_{1} e^{\frac{t}{2}} + c_{2} e^{\frac{t}{2}} t + 3 e^{\frac{t}{2}} t^{2}

y^{^{''''}} + 36 y^{^{''}} = sin (6 t) + t e^{7 t} + 4

y^{^{''}} + 8 y = 16 sin (x)

y^{^{''}} - y^{^{'}} - 2 y = 3 e^{2 x} - x^{2}

y^{^{''}} + 3 y^{^{'}} = 3

y^{^{'''}} - 2 y^{^{''}} + 2 y^{^{'}} = cos (x) + e^{x}

What is the solution for 4y^{''}-4y^'+y=24e^{t/2} ?
The solution for 4y^{''}-4y^'+y=24e^{t/2} is y=c_{1}e^{t/2}+c_{2}e^{t/2}t+3e^{t/2}t^2