What is the solution for y^{''}+2y=0,y(0)=1,y^'(0)=1 ?

The solution for y^{''}+2y=0,y(0)=1,y^'(0)=1 is y=cos(sqrt(2)t)+1/(sqrt(2))sin(sqrt(2)t)

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y^{''}+2y=0,y(0)=1,y^'(0)=1

y^{^{''}} + 2 y = 0, y (0) = 1, y^{^{'}} (0) = 1

y = cos (2 t) + \frac{1}{2} sin (2 t)

Solution steps

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

Solve linear ODE:

y = cos (2 t) + \frac{1}{2} sin (2 t)

y = cos (2 t) + \frac{1}{2} sin (2 t)

625 x^{^{''}} + 100 x^{^{'}} + 3 x = 0, x (0) = 25, x^{^{'}} (0) = - 2

y^{^{''}} + 8 y^{^{'}} + 1 y = 0, y (0) = 3, y^{^{'}} (0) = 4

2 \cdot \frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} - 3 y = 0

u^{^{''}} + 2 r u^{^{'}} (t) + a H (t) = 0

16 y^{^{''}} - y = 0, y (- 4) = 1, y^{^{'}} (- 4) = - 1

What is the solution for y^{''}+2y=0,y(0)=1,y^'(0)=1 ?
The solution for y^{''}+2y=0,y(0)=1,y^'(0)=1 is y=cos(sqrt(2)t)+1/(sqrt(2))sin(sqrt(2)t)