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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

5 y^{^{''}} - 15 y = 10 e^{x}

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

y^{^{''}} - y^{^{'}} = 16 e^{3 x}

y^{^{''}} + 2 y^{^{'}} + 2 y = 5 sin (x) - 10 x e^{x}

y^{^{''}} - 4 y^{^{'}} + 4 y = sin (x)

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