What is the solution for y^{''}+y=0,y(0)=2,y^'(0)=3 ?

The solution for y^{''}+y=0,y(0)=2,y^'(0)=3 is y=2cos(t)+3sin(t)

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y^{''}+y=0,y(0)=2,y^'(0)=3

y^{^{''}} + y = 0, y (0) = 2, y^{^{'}} (0) = 3

y = 2 cos (t) + 3 sin (t)

Solution steps

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

Solve linear ODE:

y = 2 cos (t) + 3 sin (t)

y = 2 cos (t) + 3 sin (t)

4 \frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + y = 0

41 y - 8 y^{^{'}} + y^{^{''}} = 0, y (0) = - 3, y^{^{'}} (0) = - 32

y^{^{''}} + 4 y = 0, y (0) = 3, y^{^{'}} (0) = 6

y^{^{''}} + 8 y^{^{'}} - 16 y = 0

y^{^{'''}} - 2 y^{^{''}} - 24 y^{^{'}} = 0

What is the solution for y^{''}+y=0,y(0)=2,y^'(0)=3 ?
The solution for y^{''}+y=0,y(0)=2,y^'(0)=3 is y=2cos(t)+3sin(t)