What is the solution for y^{''}+36y=0,y(0)=5,y^'(0)=-12 ?

The solution for y^{''}+36y=0,y(0)=5,y^'(0)=-12 is y=5cos(6t)-2sin(6t)

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y^{''}+36y=0,y(0)=5,y^'(0)=-12

y^{^{''}} + 36 y = 0, y (0) = 5, y^{^{'}} (0) = - 12

y = 5 cos (6 t) - 2 sin (6 t)

Solution steps

y^{''} (t) + 36 y = 0

Solve linear ODE:

y = 5 cos (6 t) - 2 sin (6 t)

y = 5 cos (6 t) - 2 sin (6 t)

y^{^{''}} + 4 \cdot y^{^{'}} + 4 \cdot y = 0

y^{^{''}} + 12 y^{^{'}} + 36 y = 0, y (0) = - 1, y^{^{'}} (0) = 9

x^{^{''}} + 3 x^{^{'}} + 3 x = 0

49 y^{^{''}} - 14 y^{^{'}} + y = 0

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

What is the solution for y^{''}+36y=0,y(0)=5,y^'(0)=-12 ?
The solution for y^{''}+36y=0,y(0)=5,y^'(0)=-12 is y=5cos(6t)-2sin(6t)