What is the solution for y^{''}+4y=0,y(0)=-5,y^'(0)=2 ?

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

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)=-5,y^'(0)=2

y^{^{''}} + 4 y = 0, y (0) = - 5, y^{^{'}} (0) = 2

y = - 5 cos (2 t) + sin (2 t)

Solution steps

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

Solve linear ODE:

y = - 5 cos (2 t) + sin (2 t)

y = - 5 cos (2 t) + sin (2 t)

y^{^{''}} + y = 0, y (\frac{π}{4}) = 0, y^{^{'}} (\frac{π}{4}) = - 3

2 y^{^{''}} + 10 y = 0, y (0) = 1, y^{^{'}} (0) = - 2

\frac{d ^{2} x}{d t ^{2}} + 3 \frac{d x}{d t} + 8 x = 0

y^{^{''''''}} + 3 y^{^{''''}} + 3 y^{^{''}} + y = 0

x^{^{''}} - 3 x = 0

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