What is the solution for y^{''}-4y^'+5y=0,y(pi/2)=3,y(0)=5 ?

The solution for y^{''}-4y^'+5y=0,y(pi/2)=3,y(0)=5 is y=e^{2t}(5cos(t)+3/(e^pi)sin(t))

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y^{''}-4y^'+5y=0,y(pi/2)=3,y(0)=5

y^{^{''}} - 4 y^{^{'}} + 5 y = 0, y (\frac{π}{2}) = 3, y (0) = 5

y = e^{2 t} (5 cos (t) + \frac{3}{e ^{π}} sin (t))

Solution steps

y^{''} (t) - 4 y^{'} (t) + 5 y = 0

Solve linear ODE:

y = e^{2 t} (5 cos (t) + \frac{3}{e ^{π}} sin (t))

y = e^{2 t} (5 cos (t) + \frac{3}{e ^{π}} sin (t))

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

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

What is the solution for y^{''}-4y^'+5y=0,y(pi/2)=3,y(0)=5 ?
The solution for y^{''}-4y^'+5y=0,y(pi/2)=3,y(0)=5 is y=e^{2t}(5cos(t)+3/(e^pi)sin(t))