What is the solution for y^{''}+6y^'+18y=0,y(pi/2)=0,y^'(pi/2)=6 ?

The solution for y^{''}+6y^'+18y=0,y(pi/2)=0,y^'(pi/2)=6 is y=2e^{(3pi)/2-3t}cos(3t)

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y^{''}+6y^'+18y=0,y(pi/2)=0,y^'(pi/2)=6

y^{^{''}} + 6 y^{^{'}} + 18 y = 0, y (\frac{π}{2}) = 0, y^{^{'}} (\frac{π}{2}) = 6

y = 2 e^{\frac{3 π}{2} - 3 t} cos (3 t)

Solution steps

y^{''} (t) + 6 y^{'} (t) + 18 y = 0

Solve linear ODE:

y = 2 e^{\frac{3 π}{2} - 3 t} cos (3 t)

y = 2 e^{\frac{3 π}{2} - 3 t} cos (3 t)

\frac{d}{d x} (5 x^{2} (x - 5)^{4})

\frac{d ^{2}}{d x ^{2}} (x^{\frac{1}{10}})

y = 2 x + 1, (1, 5)

\int cos (5 x) x d x

x \to 0 lim (\frac{ln ( x ^{2} )}{x})

What is the solution for y^{''}+6y^'+18y=0,y(pi/2)=0,y^'(pi/2)=6 ?
The solution for y^{''}+6y^'+18y=0,y(pi/2)=0,y^'(pi/2)=6 is y=2e^{(3pi)/2-3t}cos(3t)