What is the solution for x^{''}+4x=0,x(0)=8,x^'(0)=-13 ?

The solution for x^{''}+4x=0,x(0)=8,x^'(0)=-13 is x=8cos(2t)-13/2 sin(2t)

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x^{''}+4x=0,x(0)=8,x^'(0)=-13

x^{^{''}} + 4 x = 0, x (0) = 8, x^{^{'}} (0) = - 13

x = 8 cos (2 t) - \frac{13}{2} sin (2 t)

Solution steps

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

Solve linear ODE:

x = 8 cos (2 t) - \frac{13}{2} sin (2 t)

x = 8 cos (2 t) - \frac{13}{2} sin (2 t)

\frac{d ^{2} y}{d x ^{2}} = n y

4 y^{^{''}} + 9 y^{^{'}} - 9 y = 0

y^{^{''}} + 8 y^{^{'}} + 17 y = 0, y (0) = 0, y^{^{'}} (0) = 7

(d^{3} - 5 d^{2} + 4 d - 20) x = 0

\frac{y ^{^{''}}}{y} = - a^{2}

What is the solution for x^{''}+4x=0,x(0)=8,x^'(0)=-13 ?
The solution for x^{''}+4x=0,x(0)=8,x^'(0)=-13 is x=8cos(2t)-13/2 sin(2t)