What is the solution for y^{''}+1y=0,y(pi/2)=2,y^'(pi/2)=3 ?

The solution for y^{''}+1y=0,y(pi/2)=2,y^'(pi/2)=3 is y=-3cos(t)+2sin(t)

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y^{''}+1y=0,y(pi/2)=2,y^'(pi/2)=3

y^{^{''}} + 1 y = 0, y (\frac{π}{2}) = 2, y^{^{'}} (\frac{π}{2}) = 3

y = - 3 cos (t) + 2 sin (t)

Solution steps

y^{''} (t) + 1 y = 0

Solve linear ODE:

y = - 3 cos (t) + 2 sin (t)

y = - 3 cos (t) + 2 sin (t)

y^{^{''}} - 7 y^{^{'}} + 12 y = 0, y (0) = 1, y^{^{'}} (0) = \frac{5}{2}

y^{^{''}} - 6 y^{^{'}} + 9 y = 0, y (0) = 15, y (1) = 0

y^{^{''}} - 3 y^{^{'}} - 4 y = 0, y (0) = 2, y^{^{'}} (0) = 28

y^{^{''}} - 22 y^{^{'}} + 112 y = 0

y^{^{''}} + 12 y^{^{'}} + 12 y = 0

What is the solution for y^{''}+1y=0,y(pi/2)=2,y^'(pi/2)=3 ?
The solution for y^{''}+1y=0,y(pi/2)=2,y^'(pi/2)=3 is y=-3cos(t)+2sin(t)