What is the solution for 18y^{''}+72y=0,y(0)=1,y^'(0)=-6 ?

The solution for 18y^{''}+72y=0,y(0)=1,y^'(0)=-6 is y=cos(2t)-3sin(2t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

18y^{''}+72y=0,y(0)=1,y^'(0)=-6

18 y^{^{''}} + 72 y = 0, y (0) = 1, y^{^{'}} (0) = - 6

y = cos (2 t) - 3 sin (2 t)

Solution steps

18 y^{''} (t) + 72 y = 0

Solve linear ODE:

y = cos (2 t) - 3 sin (2 t)

y = cos (2 t) - 3 sin (2 t)

6 y^{^{''}} - 10 y^{^{'}} + 4 y = 0

5 y^{^{''}} + \frac{15}{0.2} y = 0, y (0) = 0, y^{^{'}} (0) = - 10

\frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} + 7 y = 0

\frac{d ^{2} y}{d t ^{2}} + 2 \frac{d y}{d t} - 15 y = 0

(4 + 8 x) y^{^{''}} - 8 y + (4 - 8 x) y = 0

What is the solution for 18y^{''}+72y=0,y(0)=1,y^'(0)=-6 ?
The solution for 18y^{''}+72y=0,y(0)=1,y^'(0)=-6 is y=cos(2t)-3sin(2t)