What is the solution for y^{'''}-5y^{''}+6y^'=6e^x ?

The solution for y^{'''}-5y^{''}+6y^'=6e^x is y=c_{1}+c_{2}e^{2x}+c_{3}e^{3x}+3e^x

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y^{'''}-5y^{''}+6y^'=6e^x

y^{^{'''}} - 5 y^{^{''}} + 6 y^{^{'}} = 6 e^{x}

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

Solution steps

y^{'''} (x) - 5 y^{''} (x) + 6 y^{'} (x) = 6 e^{x}

Solve linear ODE:

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

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

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

\frac{1}{8} x^{^{''}} + x^{^{'}} + 2 x = 12.5 \cdot sin (2 t)

y^{^{''}} + y = \frac{sin ( x )}{( cos ( x ) ) ^{2}}

\frac{d ^{2} y}{d x ^{2}} + 4 y = 1

y^{^{''}} + 9 y = 2 sin (x)

What is the solution for y^{'''}-5y^{''}+6y^'=6e^x ?
The solution for y^{'''}-5y^{''}+6y^'=6e^x is y=c_{1}+c_{2}e^{2x}+c_{3}e^{3x}+3e^x