What is the solution for y^{''}+4y=0,y(0)=5,y^'(0)=1 ?

The solution for y^{''}+4y=0,y(0)=5,y^'(0)=1 is y=5cos(2t)+1/2 sin(2t)

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

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

y = 5 cos (2 t) + \frac{1}{2} sin (2 t)

Solution steps

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

Solve linear ODE:

y = 5 cos (2 t) + \frac{1}{2} sin (2 t)

y = 5 cos (2 t) + \frac{1}{2} sin (2 t)

\int_{0}^{s} \frac{1}{0.2 s + 10} d s

\frac{\partial}{\partial x} (\frac{1}{x + 2})

h \to 0 lim (\frac{( 3 + h ) ^{3} - 9}{h})

\frac{\partial}{\partial x} (3 x^{2} - 4 x y + x yz)

y^{^{'}} = x^{2} - 2

What is the solution for y^{''}+4y=0,y(0)=5,y^'(0)=1 ?
The solution for y^{''}+4y=0,y(0)=5,y^'(0)=1 is y=5cos(2t)+1/2 sin(2t)