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

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

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

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

y = cos (t) + sin (t)

Solution steps

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

Solve linear ODE:

y = cos (t) + sin (t)

y = cos (t) + sin (t)

5 t^{3}

f (t) = 6 cos (2 t)

\frac{d}{d x} (256 cos (2 x))

\int \frac{1}{4} x^{2} - \frac{1}{2} ln (x) d x

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