What is the solution for 2y^{''}+4y=6t,y(0)=4.9,y^'(0)=0 ?

The solution for 2y^{''}+4y=6t,y(0)=4.9,y^'(0)=0 is y=4.9cos(sqrt(2)t)-1.06066…sin(sqrt(2)t)+(3t)/2

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2y^{''}+4y=6t,y(0)=4.9,y^'(0)=0

2 y^{^{''}} + 4 y = 6 t, y (0) = 4.9, y^{^{'}} (0) = 0

y = 4.9 cos (2 t) - 1.06066 \dots sin (2 t) + \frac{3 t}{2}

Solution steps

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

Solve linear ODE:

y = 4.9 cos (2 t) - 1.06066 \dots sin (2 t) + \frac{3 t}{2}

y = 4.9 cos (2 t) - 1.06066 \dots sin (2 t) + \frac{3 t}{2}

y^{^{''}} - 5 y^{^{'}} + 6 y = 24

D (t) = - 200 t + 9000

\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} + 4 y = 4

y^{^{''}} + 4 y = t^{2} + 4 e t, y (0) = 0, y^{^{'}} (0) = 3

y^{^{''}} + 9 y = 18 (sec (3 x))^{3}

What is the solution for 2y^{''}+4y=6t,y(0)=4.9,y^'(0)=0 ?
The solution for 2y^{''}+4y=6t,y(0)=4.9,y^'(0)=0 is y=4.9cos(sqrt(2)t)-1.06066…sin(sqrt(2)t)+(3t)/2