What is the solution for y^{''}+4y=9,ty(0)=0,y^'(0)=7 ?

The solution for y^{''}+4y=9,ty(0)=0,y^'(0)=7 is y=c_{1}cos(2t)+c_{2}sin(2t)+9/4

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y^{''}+4y=9,ty(0)=0,y^'(0)=7

y^{^{''}} + 4 y = 9, t y (0) = 0, y^{^{'}} (0) = 7

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^{^{''}} + 8 y^{^{'}} + 12 y = 1 - H (t)

y^{^{'''}} - 11 y^{^{''}} + 28 y^{^{'}} = 36 e^{x}, y (0) = 25

y^{^{''}} - 2 y^{^{'}} + 5 y = e x sin (x)

(D^{3} + D^{2} - 4 D - 4) y = e^{2 x} sin (3 x)

x^{^{''}} - 5 x^{^{'}} + 6 x = 2 e^{t}

What is the solution for y^{''}+4y=9,ty(0)=0,y^'(0)=7 ?
The solution for y^{''}+4y=9,ty(0)=0,y^'(0)=7 is y=c_{1}cos(2t)+c_{2}sin(2t)+9/4