What is the solution for 2y^{''}-y^'-6y=-7e^{2x} ?

The solution for 2y^{''}-y^'-6y=-7e^{2x} is y=c_{1}e^{2x}+c_{2}e^{-(3x)/2}-e^{2x}x

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2y^{''}-y^'-6y=-7e^{2x}

2 y^{^{''}} - y^{^{'}} - 6 y = - 7 e^{2 x}

y = c_{1} e^{2 x} + c_{2} e^{- \frac{3 x}{2}} - e^{2 x} x

Solution steps

2 y^{''} (x) - y^{'} (x) - 6 y = - 7 e^{2 x}

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- \frac{3 x}{2}} - e^{2 x} x

y = c_{1} e^{2 x} + c_{2} e^{- \frac{3 x}{2}} - e^{2 x} x

y^{^{''}} - 8 y^{^{'}} + 16 = 0, y (0) = 4, y (1) = 5

4 y^{^{''}} + 6 y^{^{'}} + 25 y = 4 cos (2 t)

y^{^{''}} + 8 y^{^{'}} + 6 y = 18 x^{2}

y^{^{''}} + 4 y + 5 y = 35 e^{- 4 x}, y (0) = - 6, y^{^{'}} (0) = 1

y^{^{''}} + 4 y^{^{'}} + y = 6 sin (x)

What is the solution for 2y^{''}-y^'-6y=-7e^{2x} ?
The solution for 2y^{''}-y^'-6y=-7e^{2x} is y=c_{1}e^{2x}+c_{2}e^{-(3x)/2}-e^{2x}x