What is the solution for y^{''}+4y=9,ty(0)=0,y^'(0)=0 ?

The solution for y^{''}+4y=9,ty(0)=0,y^'(0)=0 is y=c_{1}cos(2t)+c_{2}sin(2t)+9/4

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y^{''}+4y=9,ty(0)=0,y^'(0)=0

y^{^{''}} + 4 y = 9, t y (0) = 0, y^{^{'}} (0) = 0

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{9}{4}

Solution steps

y^{''} (t) + 4 y = 9

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{9}{4}

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{9}{4}

y^{^{''}} - 9 y = cos^{2} (x)

y^{^{''}} - y^{^{'}} - 6 y = e^{3 t} + 5 t

y^{^{''}} + y^{^{'}} - 3 y = 1

y^{^{''}} + 3 y^{^{'}} + 2 y = 2 cos^{2} (x)

y^{^{''}} - 24 y = 2 x cos (5 x)

What is the solution for y^{''}+4y=9,ty(0)=0,y^'(0)=0 ?
The solution for y^{''}+4y=9,ty(0)=0,y^'(0)=0 is y=c_{1}cos(2t)+c_{2}sin(2t)+9/4