What is the solution for y^{''}+3y=0,y(0)=1,y^'(0)=3 ?

The solution for y^{''}+3y=0,y(0)=1,y^'(0)=3 is y=cos(sqrt(3)t)+sqrt(3)sin(sqrt(3)t)

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

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

y = cos (3 t) + 3 sin (3 t)

Solution steps

y^{''} (t) + 3 y = 0

Solve linear ODE:

y = cos (3 t) + 3 sin (3 t)

y = cos (3 t) + 3 sin (3 t)

y^{^{''}} + 5 \cdot y = 0

y^{^{''}} - 4 y + 5 y = 0

y^{^{'''}} - 2 y^{^{''}} = 0

5 \frac{d ^{2} x}{d t ^{2}} + 3 \frac{d x}{d t} + 6 x = 0

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

What is the solution for y^{''}+3y=0,y(0)=1,y^'(0)=3 ?
The solution for y^{''}+3y=0,y(0)=1,y^'(0)=3 is y=cos(sqrt(3)t)+sqrt(3)sin(sqrt(3)t)