What is the solution for y^{''}+9y=cos(x),y(pi/2)=-5/3 ,y^'(0)=5 ?

The solution for y^{''}+9y=cos(x),y(pi/2)=-5/3 ,y^'(0)=5 is y=c_{1}cos(3x)+5/3 sin(3x)+1/8 cos(x)

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y^{''}+9y=cos(x),y(pi/2)=-5/3 ,y^'(0)=5

y^{^{''}} + 9 y = cos (x), y (\frac{π}{2}) = - \frac{5}{3}, y^{^{'}} (0) = 5

y = c_{1} cos (3 x) + \frac{5}{3} sin (3 x) + \frac{1}{8} cos (x)

Solution steps

y^{''} (x) + 9 y = cos (x)

Solve linear ODE:

y = c_{1} cos (3 x) + \frac{5}{3} sin (3 x) + \frac{1}{8} cos (x)

y = c_{1} cos (3 x) + \frac{5}{3} sin (3 x) + \frac{1}{8} cos (x)

(D^{2} - 3 D + 2) y = x + 1

4 y^{^{''}} + 4 y + y = 3 x e^{x}

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

x^{^{''}} + 8 x^{^{'}} + 16 x = e^{- t} sin (4 t)

D (t) = \frac{100 ln ( 24 + 2 )}{24 + 2}

What is the solution for y^{''}+9y=cos(x),y(pi/2)=-5/3 ,y^'(0)=5 ?
The solution for y^{''}+9y=cos(x),y(pi/2)=-5/3 ,y^'(0)=5 is y=c_{1}cos(3x)+5/3 sin(3x)+1/8 cos(x)