What is the solution for 2y^{''}+4y=6t,y(0)=0,y^'(0)=0 ?

The solution for 2y^{''}+4y=6t,y(0)=0,y^'(0)=0 is y=(3t)/2-3/(2sqrt(2))sin(sqrt(2)t)

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2y^{''}+4y=6t,y(0)=0,y^'(0)=0

2 y^{^{''}} + 4 y = 6 t, y (0) = 0, y^{^{'}} (0) = 0

y = \frac{3 t}{2} - \frac{3}{2 2} sin (2 t)

Solution steps

2 y^{''} (t) + 4 y = 6 t

Solve linear ODE:

y = \frac{3 t}{2} - \frac{3}{2 2} sin (2 t)

y = \frac{3 t}{2} - \frac{3}{2 2} sin (2 t)

y^{^{''}} + 4 y = t^{2} + 6 e^{t}, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 4 y = t^{2} + 6 e^{t}, y (0) = 0, y^{^{'}} (0) = 9

x^{^{''}} + x = 9 cos (4 t)

y^{^{''}} - 4 y = 3 e^{2 t}

9 y^{^{''}} + 7 y^{^{'}} - y = 24

What is the solution for 2y^{''}+4y=6t,y(0)=0,y^'(0)=0 ?
The solution for 2y^{''}+4y=6t,y(0)=0,y^'(0)=0 is y=(3t)/2-3/(2sqrt(2))sin(sqrt(2)t)