What is the solution for y^{''}-6y^'+34y=0,y(0)=5,y^'(0)=-5 ?

The solution for y^{''}-6y^'+34y=0,y(0)=5,y^'(0)=-5 is y=e^{3t}(5cos(5t)-4sin(5t))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-6y^'+34y=0,y(0)=5,y^'(0)=-5

y^{^{''}} - 6 y^{^{'}} + 34 y = 0, y (0) = 5, y^{^{'}} (0) = - 5

y = e^{3 t} (5 cos (5 t) - 4 sin (5 t))

Solution steps

y^{''} (t) - 6 y^{'} (t) + 34 y = 0

Solve linear ODE:

y = e^{3 t} (5 cos (5 t) - 4 sin (5 t))

y = e^{3 t} (5 cos (5 t) - 4 sin (5 t))

y^{^{''}} - 10 y^{^{'}} + 50 y = 0, y (0) = - 4, y^{^{'}} (0) = - 5

9 y^{^{''}} - 12 y^{^{'}} + 4 y = 0, y (0) = e^{(\frac{2 x}{3})}

y^{^{''}} + 2 y^{^{'}} + (1 + \frac{9}{4} π^{2}) y = 0

y^{^{''}} + 16 y = 0, y (0) = cos (4 x)

y^{^{''}} - y^{^{'}} - y = 0, y^{1} (0) = e^{x}, y^{2} (0) = x

What is the solution for y^{''}-6y^'+34y=0,y(0)=5,y^'(0)=-5 ?
The solution for y^{''}-6y^'+34y=0,y(0)=5,y^'(0)=-5 is y=e^{3t}(5cos(5t)-4sin(5t))