What is the solution for y^{''}+4y=2e^{2t},y(0)=1,y^'(0)=0 ?

The solution for y^{''}+4y=2e^{2t},y(0)=1,y^'(0)=0 is y= 3/4 cos(2t)-1/4 sin(2t)+1/4 e^{2t}

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y^{''}+4y=2e^{2t},y(0)=1,y^'(0)=0

y^{^{''}} + 4 y = 2 e^{2 t}, y (0) = 1, y^{^{'}} (0) = 0

y = \frac{3}{4} cos (2 t) - \frac{1}{4} sin (2 t) + \frac{1}{4} e^{2 t}

Solution steps

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

Solve linear ODE:

y = \frac{3}{4} cos (2 t) - \frac{1}{4} sin (2 t) + \frac{1}{4} e^{2 t}

y = \frac{3}{4} cos (2 t) - \frac{1}{4} sin (2 t) + \frac{1}{4} e^{2 t}

y^{^{''}} - y^{^{'}} - 6 y = 2 sin (t)

D (x) = 105.6 - 2.4 x

x^{^{''}} + 4 x^{^{'}} + 43 x = 10 cos (6 t)

y^{^{''}} - 3 y^{^{''}} + 3 y^{^{'}} - y = x - 2 e^{x}

y^{^{''}} + 4 y^{^{'}} + 5 y = 1 - 5 x + 8 cos (x) - 8 sin (x)

What is the solution for y^{''}+4y=2e^{2t},y(0)=1,y^'(0)=0 ?
The solution for y^{''}+4y=2e^{2t},y(0)=1,y^'(0)=0 is y= 3/4 cos(2t)-1/4 sin(2t)+1/4 e^{2t}