What is the solution for y^{''}+9y=0,y(pi/2)=1,y^'(pi/2)=-3 ?

The solution for y^{''}+9y=0,y(pi/2)=1,y^'(pi/2)=-3 is y=-cos(3t)-sin(3t)

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y^{''}+9y=0,y(pi/2)=1,y^'(pi/2)=-3

y^{^{''}} + 9 y = 0, y (\frac{π}{2}) = 1, y^{^{'}} (\frac{π}{2}) = - 3

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

Solution steps

y^{''} (t) + 9 y = 0

Solve linear ODE:

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

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

\frac{d y}{d x}, 9 y + x = 4 y

\int (6 x^{2} - 4 e^{x}) d x

\frac{5 x ^{3} + 5 x ^{2}}{x}

\int \frac{arctan ( x )}{x ^{2}} d x

x = 4 y^{2}, x + y = 3

What is the solution for y^{''}+9y=0,y(pi/2)=1,y^'(pi/2)=-3 ?
The solution for y^{''}+9y=0,y(pi/2)=1,y^'(pi/2)=-3 is y=-cos(3t)-sin(3t)