What is the solution for y^{''}+4y=9t,y^'(0)=7 ?

The solution for y^{''}+4y=9t,y^'(0)=7 is y=c_{1}cos(2t)+19/8 sin(2t)+(9t)/4

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

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

y = c_{1} cos (2 t) + \frac{19}{8} sin (2 t) + \frac{9 t}{4}

Solution steps

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

Solve linear ODE:

y = c_{1} cos (2 t) + \frac{19}{8} sin (2 t) + \frac{9 t}{4}

y = c_{1} cos (2 t) + \frac{19}{8} sin (2 t) + \frac{9 t}{4}

y^{^{''}} + 3 y^{^{'}} - 4 y = 2 cos (3 x)

y^{^{''}} + 2 y^{^{'}} + y = t^{2} - 6

y^{^{''}} + 2 y^{^{'}} + 2 y = \frac{e ^{- x}}{cos ( x )}

y^{^{''}} - y = ln (t)

y^{^{''}} - 12 y^{^{'}} + 36 y = 12 x + 5

What is the solution for y^{''}+4y=9t,y^'(0)=7 ?
The solution for y^{''}+4y=9t,y^'(0)=7 is y=c_{1}cos(2t)+19/8 sin(2t)+(9t)/4