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Problems of second-order differential equations in higher mathematics
Because 3x+ 1 = (3x+ 1) e (0 * x)

For y "+py'+q = f (x) e (enter *x)

Enter =0 is not a characteristic root, so let the special solution be

y=ax+b

Y'=a y"=0 into the original equation.

0-2a-2(ax+b)=3x+ 1

-2ax-2(a+b)=3x+ 1

-2a=3 -2(a+b)= 1

a=-3/2

b= 1

Special solution y=-3x/2+ 1

The general solution y = c1e (3x)+c2e (-x)-3x/2+1.

For f (x) e (converted to x),

f(x)=a 1x^(nx)+a2x^[(n- 1)x]+...+a(n+ 1)

When in is one of the characteristic roots, let the special solution be

Y = xq (x) e (enter x) q (x) = a1x (NX)+...+a (n+1)

If it is not a characteristic root, let y = q (x) e (into x)