What is the solution for y^{''}+4y=0,y(0)=1,y(1)=0 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+4y=0,y(0)=1,y(1)=0

y^{^{''}} + 4 y = 0, y (0) = 1, y (1) = 0

y = cos (2 t) - \frac{cos ( 2 ) sin ( 2 t )}{sin ( 2 )}

Solution steps

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

Solve linear ODE:

y = cos (2 t) - \frac{cos ( 2 ) sin ( 2 t )}{sin ( 2 )}

y = cos (2 t) - \frac{cos ( 2 ) sin ( 2 t )}{sin ( 2 )}

\frac{d}{d t} (a e^{t} cos (2 t) + b e^{t} sin (2 t))

x \to 0 lim ((1 + x)^{2} - 1)

x \to 1 lim (6 x^{2} + 1)

\int \frac{1}{4 + v ^{2}} d v

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