What is the solution for ty^'-y=3t^2cos(2t) ?

The solution for ty^'-y=3t^2cos(2t) is y= 3/2 sin(2t)t+c_{1}t

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

ty^'-y=3t^2cos(2t)

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

y = \frac{3}{2} sin (2 t) t + c_{1} t

Solution steps

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

Solve linear ODE:

y = \frac{3}{2} sin (2 t) t + c_{1} t

y = \frac{3}{2} sin (2 t) t + c_{1} t

\int x (8 x + 5)^{8} d x

\int x - \frac{1}{2} d x

y^{^{''}} + 3 y^{^{'}} + 2 y = \frac{1}{2 + e ^{x}}

f (x) = (1 + 4 x)^{\frac{3}{2}}, at x = 2

y = x^{2} - 5 x - 3

What is the solution for ty^'-y=3t^2cos(2t) ?
The solution for ty^'-y=3t^2cos(2t) is y= 3/2 sin(2t)t+c_{1}t