What is the solution for 2y^{''}+4y=6t,y(0)=4.9,y^'(0)=0 ?

The solution for 2y^{''}+4y=6t,y(0)=4.9,y^'(0)=0 is y=4.9cos(sqrt(2)t)-1.06066…sin(sqrt(2)t)+(3t)/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

2y^{''}+4y=6t,y(0)=4.9,y^'(0)=0

2 y^{^{''}} + 4 y = 6 t, y (0) = 4.9, y^{^{'}} (0) = 0

y = 4.9 cos (2 t) - 1.06066 \dots sin (2 t) + \frac{3 t}{2}

Solution steps

2 y^{''} (t) + 4 y = 6 t

Solve linear ODE:

y = 4.9 cos (2 t) - 1.06066 \dots sin (2 t) + \frac{3 t}{2}

y = 4.9 cos (2 t) - 1.06066 \dots sin (2 t) + \frac{3 t}{2}

What is the solution for 2y^{''}+4y=6t,y(0)=4.9,y^'(0)=0 ?
The solution for 2y^{''}+4y=6t,y(0)=4.9,y^'(0)=0 is y=4.9cos(sqrt(2)t)-1.06066…sin(sqrt(2)t)+(3t)/2