What is the solution for y^{''}+4y=sin(3t),y(0)=0,y^'(0)=0 ?

The solution for y^{''}+4y=sin(3t),y(0)=0,y^'(0)=0 is y= 3/10 sin(2t)-1/5 sin(3t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+4y=sin(3t),y(0)=0,y^'(0)=0

y^{^{''}} + 4 y = sin (3 t), y (0) = 0, y^{^{'}} (0) = 0

y = \frac{3}{10} sin (2 t) - \frac{1}{5} sin (3 t)

Solution steps

y^{''} (t) + 4 y = sin (3 t)

Solve linear ODE:

y = \frac{3}{10} sin (2 t) - \frac{1}{5} sin (3 t)

y = \frac{3}{10} sin (2 t) - \frac{1}{5} sin (3 t)

y^{^{''}} + 6 y^{^{'}} - 7 y = e^{- x}

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

f (x) + f^{^{'}} (x) + f^{^{''}} (x) = x^{5} + 64

y^{^{''''}} + 9 y^{^{''}} = sin (3 t) + t e^{4 t} + 4

\frac{d ^{2} y}{d x ^{2}} - \frac{d y}{d x} - 2 y = 4 x^{2}

What is the solution for y^{''}+4y=sin(3t),y(0)=0,y^'(0)=0 ?
The solution for y^{''}+4y=sin(3t),y(0)=0,y^'(0)=0 is y= 3/10 sin(2t)-1/5 sin(3t)