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

The solution for y^{''}-4y^'+3y=2x+1 is y=c_{1}e^{3x}+c_{2}e^x+(2x)/3+11/9

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y^{''}-4y^'+3y=2x+1

y^{^{''}} - 4 y^{^{'}} + 3 y = 2 x + 1

y = c_{1} e^{3 x} + c_{2} e^{x} + \frac{2 x}{3} + \frac{11}{9}

Solution steps

y^{''} (x) - 4 y^{'} (x) + 3 y = 2 x + 1

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{x} + \frac{2 x}{3} + \frac{11}{9}

y = c_{1} e^{3 x} + c_{2} e^{x} + \frac{2 x}{3} + \frac{11}{9}

\frac{d ^{2} y}{d x ^{2}} - 7 \frac{d y}{d x} = - 3

y^{^{''}} + y^{^{'}} = 2 sin (2) t

y^{^{''}} - 6 y^{^{'}} + 25 y = 30 e^{3 x} sin (x)

y^{^{''}} + y^{^{'}} - 2 y = x + sin (2 x)

y^{^{''}} + 2 y^{^{'}} + 10 y = 2 cos (6 x)

What is the solution for y^{''}-4y^'+3y=2x+1 ?
The solution for y^{''}-4y^'+3y=2x+1 is y=c_{1}e^{3x}+c_{2}e^x+(2x)/3+11/9