What is the solution for y^{''}-2y^'+10y=0,y(0)=-2,y^'(0)=5 ?

The solution for y^{''}-2y^'+10y=0,y(0)=-2,y^'(0)=5 is y=e^t(-2cos(3t)+7/3 sin(3t))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}-2y^'+10y=0,y(0)=-2,y^'(0)=5

y^{^{''}} - 2 y^{^{'}} + 10 y = 0, y (0) = - 2, y^{^{'}} (0) = 5

y = e^{t} (- 2 cos (3 t) + \frac{7}{3} sin (3 t))

Solution steps

y^{''} (t) - 2 y^{'} (t) + 10 y = 0

Solve linear ODE:

y = e^{t} (- 2 cos (3 t) + \frac{7}{3} sin (3 t))

y = e^{t} (- 2 cos (3 t) + \frac{7}{3} sin (3 t))

\frac{d}{d x} ((8 - 3 x^{2})^{3})

y^{^{'}} = x ln (x), y (1) = 2

\int x (2 + x^{3})^{2} d x

\int I n^{2} d x

\int \frac{1 + cos ( x )}{cos ^{2} ( x )} d x

What is the solution for y^{''}-2y^'+10y=0,y(0)=-2,y^'(0)=5 ?
The solution for y^{''}-2y^'+10y=0,y(0)=-2,y^'(0)=5 is y=e^t(-2cos(3t)+7/3 sin(3t))