What is the solution for y^{''}+4y=0,y(0)=1,y^'(0)=2 ?

The solution for y^{''}+4y=0,y(0)=1,y^'(0)=2 is y=cos(2t)+sin(2t)

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y^{''}+4y=0,y(0)=1,y^'(0)=2

y^{^{''}} + 4 y = 0, y (0) = 1, y^{^{'}} (0) = 2

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

Solution steps

y^{''} (t) + 4 y = 0

Solve linear ODE:

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

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

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

y^{^{''}} + 2 y^{^{'}} + y = 0, y (0) = 10, y^{^{'}} (0) = - 9

y^{^{''}} - 4 y^{^{'}} - 4 y = 0, y (0) = 1, y^{^{'}} (0) = 5

y^{^{''}} - 4 y^{^{'}} - 4 y = 0, y (0) = 2, y^{^{'}} (0) = 16

y^{^{''}} + 2 y^{^{'}} - 8 y = 0, y (0) = 3, y (0) = - 12

What is the solution for y^{''}+4y=0,y(0)=1,y^'(0)=2 ?
The solution for y^{''}+4y=0,y(0)=1,y^'(0)=2 is y=cos(2t)+sin(2t)