What is the solution for y^{''}+y=0,y(0)=19,y^'(0)=36 ?

The solution for y^{''}+y=0,y(0)=19,y^'(0)=36 is y=19cos(t)+36sin(t)

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y^{''}+y=0,y(0)=19,y^'(0)=36

y^{^{''}} + y = 0, y (0) = 19, y^{^{'}} (0) = 36

y = 19 cos (t) + 36 sin (t)

Solution steps

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

Solve linear ODE:

y = 19 cos (t) + 36 sin (t)

y = 19 cos (t) + 36 sin (t)

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

y^{^{''}} + a^{2} \cdot n^{2} \cdot π^{2} \cdot y = 0

3 y^{^{''}} - 4 y^{^{'}} = 0

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + 29 y = 0

y^{^{'''}} - 5 y^{^{''}} + 3 y^{^{'}} + y = 0

What is the solution for y^{''}+y=0,y(0)=19,y^'(0)=36 ?
The solution for y^{''}+y=0,y(0)=19,y^'(0)=36 is y=19cos(t)+36sin(t)