What is the solution for 2y^{''}-3y^'+y=64 ?

The solution for 2y^{''}-3y^'+y=64 is y=c_{1}e^t+c_{2}e^{t/2}+64

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2y^{''}-3y^'+y=64

2 y^{^{''}} - 3 y^{^{'}} + y = 64

y = c_{1} e^{t} + c_{2} e^{\frac{t}{2}} + 64

Solution steps

2 y^{''} (t) - 3 y^{'} (t) + y = 64

Solve linear ODE:

y = c_{1} e^{t} + c_{2} e^{\frac{t}{2}} + 64

y = c_{1} e^{t} + c_{2} e^{\frac{t}{2}} + 64

y^{^{''}} - 3 y + 2 = x^{2} + 2 x + 1

y^{^{''}} - y^{^{'}} = x^{2} + 1

5 y^{^{''}} + 7 y^{^{'}} + 3 y = x^{2} + 2 e^{5 x}

(D^{2} - 2 D + 1) y = cosh (x)

x^{^{''}} + a^{2} x = a^{2} b - c

What is the solution for 2y^{''}-3y^'+y=64 ?
The solution for 2y^{''}-3y^'+y=64 is y=c_{1}e^t+c_{2}e^{t/2}+64