What is the solution for y^{''}-4y^'+3y=3e^{2x} ?

The solution for y^{''}-4y^'+3y=3e^{2x} is y=c_{1}e^{3x}+c_{2}e^x-3e^{2x}

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y^{''}-4y^'+3y=3e^{2x}

y^{^{''}} - 4 y^{^{'}} + 3 y = 3 e^{2 x}

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

Solution steps

y^{''} (x) - 4 y^{'} (x) + 3 y = 3 e^{2 x}

Solve linear ODE:

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

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

y^{^{''}} - 7 y^{^{'}} = 5 x \cdot e^{x}

y^{^{''}} + y^{^{'}} - 32 y = 16 cos (x)

y^{^{''}} + 4 y = 32 sin (4 t), y (0) = 1

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

y^{^{''}} - 12 y^{^{'}} + 5 y = - 8 sin (t)

What is the solution for y^{''}-4y^'+3y=3e^{2x} ?
The solution for y^{''}-4y^'+3y=3e^{2x} is y=c_{1}e^{3x}+c_{2}e^x-3e^{2x}