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How to do the sixth sub-question of the fifth edition of Tongji Higher Mathematics on page 3 17 1?
You should mean volume two.

The corresponding odd equation y''-6y'+9y=0.

Its characteristic equation is r 2-6r+9 = 0.

The characteristic root of the solution is r=3 (triple)

So the general solution of odd equation is y * = (c 1+c2x) e 3x.

Let a special solution of the original equation be y0 = x 2 (ax+b) e 3x.

Substituting into the original equation, A= 1/6 B= 1/2.

Therefore, y0 = x 2 (1/6x+1/2) e 3x.

So the general solution is y = (c1+c2x) e3x+x2/2 (x/3+1) e3x.