What is the solution for 2500y^{''}+729y=0,y(0)=2,y^'(0)=3 ?

The solution for 2500y^{''}+729y=0,y(0)=2,y^'(0)=3 is y=2cos((27t)/(50))+50/9 sin((27t)/(50))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2500y^{''}+729y=0,y(0)=2,y^'(0)=3

2500 y^{^{''}} + 729 y = 0, y (0) = 2, y^{^{'}} (0) = 3

y = 2 cos (\frac{27 t}{50}) + \frac{50}{9} sin (\frac{27 t}{50})

Solution steps

2500 y^{''} (t) + 729 y = 0

Solve linear ODE:

y = 2 cos (\frac{27 t}{50}) + \frac{50}{9} sin (\frac{27 t}{50})

y = 2 cos (\frac{27 t}{50}) + \frac{50}{9} sin (\frac{27 t}{50})

x^{^{''}} + 8 x^{^{'}} + 10 x = 0, x (0) = 10, x^{^{'}} (0) = 0

2500 y^{^{''}} + 729 y = 0, y (0) = 2, y^{^{'}} (0) = 5

35 y^{^{''''}} + 52 y^{^{''}} + 9 y = 0

(D^{4} + D^{2}) y = 0

\frac{d ^{4} y}{d t ^{4}} - y^{^{'''}} - y^{^{''}} + y^{^{'}} = 0

What is the solution for 2500y^{''}+729y=0,y(0)=2,y^'(0)=3 ?
The solution for 2500y^{''}+729y=0,y(0)=2,y^'(0)=3 is y=2cos((27t)/(50))+50/9 sin((27t)/(50))