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

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

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y^{''}+4y=0,y(pi/4)=-2

y^{^{''}} + 4 y = 0, y (\frac{π}{4}) = - 2

y = c_{1} cos (2 t) - 2 sin (2 t)

Solution steps

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

Solve linear ODE:

y = c_{1} cos (2 t) - 2 sin (2 t)

y = c_{1} cos (2 t) - 2 sin (2 t)

x^{^{''}} + \frac{1}{4} x^{^{'}} + x = 0

\frac{d ^{2} x}{d t ^{2}} = - k \frac{d x}{d t}

x^{^{''}} + \frac{32}{2} x = 0, x (0) = - 1, x^{^{'}} (0) = - 2

x^{^{''}} + \frac{32}{2} x = 0, x (0) = - 1, x^{^{'}} (0) = - 5

\frac{d ^{2} y}{d x ^{2}} + 9 \frac{d y}{d x} + 8 y = 0

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