What is the solution for (y^{''}-4y^'+3y)=2x+5 ?

The solution for (y^{''}-4y^'+3y)=2x+5 is y=c_{1}e^{3x}+c_{2}e^x+(2x)/3+23/9

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(y^{''}-4y^'+3y)=2x+5

(y^{^{''}} - 4 y^{^{'}} + 3 y) = 2 x + 5

y = c_{1} e^{3 x} + c_{2} e^{x} + \frac{2 x}{3} + \frac{23}{9}

Solution steps

(y^{''} (x) - 4 y^{'} (x) + 3 y) = 2 x + 5

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{x} + \frac{2 x}{3} + \frac{23}{9}

y = c_{1} e^{3 x} + c_{2} e^{x} + \frac{2 x}{3} + \frac{23}{9}

(d^{2} + 3 d + 2) y = (e^{e})^{x}

y^{^{''}} + 6 y^{^{'}} + 9 y = \frac{e ^{- 3} x}{x}

(d^{3} + d^{2} - 4 d - 4) y = 3 e^{- x} - 4 x - 6

(d^{2} - 2 d + 4) \cdot y = e^{x} \cdot cos (x)

y^{^{'''}} - 3 y^{^{''}} - 10 y^{^{'}} + 24 = 2 x^{2} - 5

What is the solution for (y^{''}-4y^'+3y)=2x+5 ?
The solution for (y^{''}-4y^'+3y)=2x+5 is y=c_{1}e^{3x}+c_{2}e^x+(2x)/3+23/9