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

The solution for y^{''}-4y^'+3y=4e^3x is y=c_{1}e^{3x}+c_{2}e^x+(4e^3x)/3+(16e^3)/9

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

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

y = c_{1} e^{3 x} + c_{2} e^{x} + \frac{4 e ^{3} x}{3} + \frac{16 e ^{3}}{9}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{x} + \frac{4 e ^{3} x}{3} + \frac{16 e ^{3}}{9}

y = c_{1} e^{3 x} + c_{2} e^{x} + \frac{4 e ^{3} x}{3} + \frac{16 e ^{3}}{9}

(d^{4} - 1) y = e^{x} cos (x)

(d^{2} + 1) y = sec (x) + tan (x)

y^{^{''}} - 2 y^{^{'}} = x^{2} - x

(d^{2} + 9) y = 7 e^{3 x} + 3 x^{2} + 4

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

What is the solution for y^{''}-4y^'+3y=4e^3x ?
The solution for y^{''}-4y^'+3y=4e^3x is y=c_{1}e^{3x}+c_{2}e^x+(4e^3x)/3+(16e^3)/9