What is the solution for y^{''}+5y=0,y(0)=1,y^'(0)=1 ?

The solution for y^{''}+5y=0,y(0)=1,y^'(0)=1 is y=cos(sqrt(5)t)+1/(sqrt(5))sin(sqrt(5)t)

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y^{''}+5y=0,y(0)=1,y^'(0)=1

y^{^{''}} + 5 y = 0, y (0) = 1, y^{^{'}} (0) = 1

y = cos (5 t) + \frac{1}{5} sin (5 t)

Solution steps

y^{''} (t) + 5 y = 0

Solve linear ODE:

y = cos (5 t) + \frac{1}{5} sin (5 t)

y = cos (5 t) + \frac{1}{5} sin (5 t)

\frac{\partial}{\partial x} (cos (2 y - x))

\frac{d}{d x} (lo g_{10} (1 - 2 x^{2}))

(1 + ln (x) + \frac{y}{x}) d x = (6 - ln (x)) d y

\frac{x d y}{d x} - \frac{2 d y}{d x} - (x - 2) sin (x) + y = 0

\frac{\partial}{\partial x} (4 x e^{x})

What is the solution for y^{''}+5y=0,y(0)=1,y^'(0)=1 ?
The solution for y^{''}+5y=0,y(0)=1,y^'(0)=1 is y=cos(sqrt(5)t)+1/(sqrt(5))sin(sqrt(5)t)