What is the solution for y^{''''}-42y^{'''}+441y^{''}=0 ?

The solution for y^{''''}-42y^{'''}+441y^{''}=0 is y=c_{1}+c_{2}t+c_{3}e^{21t}+c_{4}te^{21t}

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y^{''''}-42y^{'''}+441y^{''}=0

y^{^{''''}} - 42 y^{^{'''}} + 441 y^{^{''}} = 0

y = c_{1} + c_{2} t + c_{3} e^{21 t} + c_{4} t e^{21 t}

Solution steps

y^{(4)} (t) - 42 y^{'''} (t) + 441 y^{''} (t) = 0

Solve linear ODE:

y = c_{1} + c_{2} t + c_{3} e^{21 t} + c_{4} t e^{21 t}

y = c_{1} + c_{2} t + c_{3} e^{21 t} + c_{4} t e^{21 t}

{\frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} + 20 y}^{2} = 0

3 y^{^{'''}} + 17 y^{^{''}} + 31 y^{^{'}} + 14 y = 0

2 y^{^{''}} - 3 y^{^{'}} + \frac{1}{2} y = 0

\frac{d ^{4} y}{d x ^{4}} - 2 \frac{d ^{2} y}{d x ^{2}} + y = 0

\frac{d ^{2} f}{d x ^{2}} = a^{2} f (x)

What is the solution for y^{''''}-42y^{'''}+441y^{''}=0 ?
The solution for y^{''''}-42y^{'''}+441y^{''}=0 is y=c_{1}+c_{2}t+c_{3}e^{21t}+c_{4}te^{21t}