Q10E

Question

find a general solution to the given equation.

$y''' + 4 y'' + y' - 26 y = e^{- 3 x} s i n 2 x + x$

Step-by-Step Solution

Verified

Answer

$y (x) = c_{3} e^{2 x} + c_{4} e^{- 3 x} \cos 2 x + c_{6} e^{- 3 x} \sin 2 x - \frac{1}{676} - \frac{x}{26} + \frac{1}{29} x e^{- 3 x} \cos 2 x - \frac{1}{58} x e^{- 3 x} \sin 2 x$

1Step 1: Find the corresponding auxiliary equation

The auxiliary equation of the corresponding homogeneous equation:

$r^{2} (r - 2) {(r^{2} + 6 r + 13)}^{2} = 0$

The solutions of the auxiliary equation are:

$r_{1} = r_{2} = 0, r_{3} = 2, r_{4} = r_{5} = - 3 + 2 i, r_{6} = r_{7} = - 3 - 2 i$

Therefore a general solution to the homogeneous equation is:

$y_{h} (x) = c_{3} e^{2 x} c_{4} e^{- 3 x} \cos 2 x + c_{6} e^{- 3 x} \sin 2 x$

2Step 2: Find the particular solution

Let the particular solution be;

$y_{p} (x) = c_{1} + c_{2} x + c_{5} x e^{- 3 x} \cos 2 x + c_{7} x e^{- 3 x} \sin 2 x$

Then,

$y_{p}^{'} (x) = c_{2} + c_{5} e^{- 3 x} \cos 2 x + c_{7} e^{- 3 x} \sin 2 x + (- 3 c_{5} + 2 c_{7}) x e^{- 3 x} \cos 2 x + (- 2 c_{5} - 3 c_{7}) x e^{- 3 x} \sin 2 x$

$y_{p}^{''} (x) = (- 6 c_{5} + 4 c_{7}) e^{- 3 x} \cos 2 x + (- 4 c_{5} - 6 c_{7}) e^{- 3 x} \sin 2 x + (5 c_{5} - 12 c_{7}) x e^{- 3 x} \cos 2 x + (12 c_{5} + 5 c_{7}) x e^{- 3 x} \sin 2 x$

$y_{p}^{'''} (x) = (15 c_{5} - 36 c_{7}) e^{- 3 x} \cos 2 x + (36 c_{5} + 15 c_{7}) e^{- 3 x} \sin 2 x + (9 c_{5} + 46 c_{7}) x e^{- 3 x} \cos 2 x + (- 46 c_{5} + 9 c_{7}) x e^{- 3 x} \sin 2 x$

Then,

$y_{p}^{'''} + 4 y_{p}^{''} + y_{p}^{'} - 36 y_{p} = (- 8 c_{5} - 20 c_{7}) e^{- 3 x} \cos 2 x + (20 c_{5} - 8 c_{7}) e^{- 3 x} \sin 2 x + c_{2} - 26 c_{1} - 26 c_{2} x = e^{- 3 x} \sin 2 x + x$

Then we have:

$- 8 c_{5} - 20 c_{7} = 0 c_{2} - 26 c_{1} = 0 20 c_{5} + 8 c_{7} = 1 - 26 c_{2} = 1$

Therefore,

$c_{1} = - \frac{1}{676}, c_{2} = - \frac{1}{26}, c_{5} = \frac{1}{29}, c_{7} = - \frac{1}{58}$

Hence,

$y_{p} (x) = - \frac{1}{676} - \frac{1}{26} x + \frac{1}{29} x e^{- 3 x} \cos 2 x - \frac{1}{58} x e^{- 3 x} \sin 2 x .$

3Step 3: y ( x ) = y h + y p

Thus, the required general solution is:

$y (x) = c_{3} e^{2 x} + c_{4} e^{- 3 x} \cos 2 x + c_{6} e^{- 3 x} \sin 2 x - \frac{1}{676} - \frac{x}{26} + \frac{1}{29} x e^{- 3 x} \cos 2 x - \frac{1}{58} x e^{- 3 x} \sin 2 x$

Q9E

Q11E

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Q11E

. find a differential operator that annihilates the given function.x4-x2+11

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Q12E

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