What is the solution for x^{'''}-10x^{''}+37x^'-52x=0 ?

The solution for x^{'''}-10x^{''}+37x^'-52x=0 is x=c_{1}e^{4t}+e^{3t}(c_{2}cos(2t)+c_{3}sin(2t))

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x^{'''}-10x^{''}+37x^'-52x=0

x^{^{'''}} - 10 x^{^{''}} + 37 x^{^{'}} - 52 x = 0

x = c_{1} e^{4 t} + e^{3 t} (c_{2} cos (2 t) + c_{3} sin (2 t))

Solution steps

x^{'''} (t) - 10 x^{''} (t) + 37 x^{'} (t) - 52 x = 0

Solve linear ODE:

x = c_{1} e^{4 t} + e^{3 t} (c_{2} cos (2 t) + c_{3} sin (2 t))

x = c_{1} e^{4 t} + e^{3 t} (c_{2} cos (2 t) + c_{3} sin (2 t))

x^{^{''}} + 6 x^{^{'}} + 8 x = 0, x (0) = 7

y^{^{''}} + 8 y^{^{'}} + 32 y = 0, y^{^{'}} 0 = - 4, y^{0} = 1

4 y^{^{''}} + 3 y^{^{''}} + y = 0

2 y^{^{'}} + 3 y^{^{''}} = 0

4 y^{^{''}} - 4 y^{^{'}} + (- 3) y = 0, y (0) = 5, y^{^{'}} (0) = 0

What is the solution for x^{'''}-10x^{''}+37x^'-52x=0 ?
The solution for x^{'''}-10x^{''}+37x^'-52x=0 is x=c_{1}e^{4t}+e^{3t}(c_{2}cos(2t)+c_{3}sin(2t))