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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

- 2 y^{^{''}} - 10 y^{^{'}} + \frac{5}{2} y = 0

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

x^{^{'''}} + x^{^{''}} - 6 x^{^{'}} = 0

4 y^{^{''}} - y^{^{'}} + 17 y = 0

y^{^{''}} + \frac{4}{3} y^{^{'}} + y = 0, y (0) = 1, y^{^{'}} (0) = - \frac{10}{3}

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