What is the solution for y^{''}+36y=0,y(0)=4,y^'(0)=-4 ?

The solution for y^{''}+36y=0,y(0)=4,y^'(0)=-4 is y=4cos(6t)-2/3 sin(6t)

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y^{''}+36y=0,y(0)=4,y^'(0)=-4

y^{^{''}} + 36 y = 0, y (0) = 4, y^{^{'}} (0) = - 4

y = 4 cos (6 t) - \frac{2}{3} sin (6 t)

Solution steps

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

Solve linear ODE:

y = 4 cos (6 t) - \frac{2}{3} sin (6 t)

y = 4 cos (6 t) - \frac{2}{3} sin (6 t)

y^{^{''''''}} - 64 y = 0

\frac{d ^{2} y}{d x ^{2}} + 3 \frac{d y}{d x} = 0

x^{^{''}} - x^{^{'}} - 1 = - 1

y^{^{''}} - y^{^{'}} - 30 y = 0, y (0) = 0, y^{^{'}} (0) = 5

49 y^{^{'''}} + 28 y^{^{''}} + 4 y^{^{'}} = 0

What is the solution for y^{''}+36y=0,y(0)=4,y^'(0)=-4 ?
The solution for y^{''}+36y=0,y(0)=4,y^'(0)=-4 is y=4cos(6t)-2/3 sin(6t)