What is the solution for 3y^{''}-5y^'-2y=3e^{2x} ?

The solution for 3y^{''}-5y^'-2y=3e^{2x} is y=c_{1}e^{2x}+c_{2}e^{-x/3}+(3e^{2x}x)/7

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3y^{''}-5y^'-2y=3e^{2x}

3 y^{^{''}} - 5 y^{^{'}} - 2 y = 3 e^{2 x}

y = c_{1} e^{2 x} + c_{2} e^{- \frac{x}{3}} + \frac{3 e ^{2 x} x}{7}

Solution steps

3 y^{''} (x) - 5 y^{'} (x) - 2 y = 3 e^{2 x}

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- \frac{x}{3}} + \frac{3 e ^{2 x} x}{7}

y = c_{1} e^{2 x} + c_{2} e^{- \frac{x}{3}} + \frac{3 e ^{2 x} x}{7}

4 y^{^{''}} - 8 y^{^{'}} = cos (2 x) + 7 x^{2}

y^{^{''}} - 2 y^{^{'}} + y = \frac{4 e ^{x}}{x ^{2} + 1}

y^{^{''}} + y^{^{'}} = x + e^{x} + x^{4}

\frac{d ^{2} y}{d x ^{2}} + y = csc (x) cot (x)

y^{^{''}} - 4 y^{^{'}} + 3 = e^{- 5 x}, y (0) = 0, y^{^{'}} (0) = 0

What is the solution for 3y^{''}-5y^'-2y=3e^{2x} ?
The solution for 3y^{''}-5y^'-2y=3e^{2x} is y=c_{1}e^{2x}+c_{2}e^{-x/3}+(3e^{2x}x)/7