4. If you choose D, you have it.

y ”+ Acos2x+Bsin2x = cos2x，

y″=( 1-A)cos2x-Bsin2x

∫y = Acos2x+bsin 2x，

∴y′=-2asin2x+2bcos2x

y″=-4Acos2x-4Bsin2x

The corresponding coefficients are equal, (1-a) =-4a, and a =-1/3;

-B=-4B，B=0，

A special solution is: y =-( 1/3) cos2x.

Therefore, y'=(2/3)sin 2x.

y″=(4/3)cos2x

y "+ y = y "-( 1/3)cos2x =(4/3)cos2x-( 1/3)cos2x = cos2x

y ”+ y = cos2x

Whether the restore verification is correct,

6, y "-2y'+y”-2y '+5y =(e x)cos2x, the special solution form is y = (e x) y = (e x) (acos2x+bsin2x), and its principle is similar to that of the four questions, all of which are undetermined coefficient methods, and the option is B.

y=(e^x)(acos2x+bsin2x)=a(e^x)cos2x+b(e^x)sin2x，

5y= 5A(e^x)cos2x+ 5B(e^x)sin2x

y′= a[(e^x)cos2x-2(e^x)sin2x]+b[(e^x)sin2x+2(e^x)cos2x]

=(a+2b)(e^x)cos2x+(-2a+ b)(e^x)sin2x

y″=(a+2b)[(e^x)cos2x-2(e^x)sin2x]+(-2a+ b)[(e^x)sin2x+2(e^x)cos2x]

=[(a+2b)+2(-2a+b)](e^x)cos2x+[-2(a+2b)+(-2a+ b)]

=(-3a+4b)(e^x)cos2x-(4a+3b)(e^x)sin2x

y″-2y′+5y=[(-3a+4b)(e^x)cos2x-(4a+3)(e^x)sin2x]

-2[(a+2b)(e^x)cos2x+(-2a+ b)(e^x)sin2x]

+[5A(e^x)cos2x+ 5B(e^x)sin2x]

=[(-3a+4b)-2(a+2b)+5](e^x)cos2x

+[-(4a+3b)-2(-2a+b)+5](e^x)sin2x

=(-5a+5)(e^x)cos2x+(-5b+5)(e^x)sin2x=(e^x)sin2x

The corresponding coefficients are equal, (-5a+5) = 1, a = 4/5; (-5B+5)=0，B= 1

y=(e^x)(acos2x+bsin2x)=(4/5)(e^x)cos2x+(e^x)sin2x，

7、y″-4y′+4y = 6x？ +8e^2x

The form of the special solution is: y=Ax? +Bx+C+Dx？ E^2x, choose b, C=0,

The corresponding coefficients are equal, 4A=6, a = 3/2; (2A-4B)=0，A=2B，B = 3/4； 2D=8，D=4 .

The solution is A=3/2, A=2B, b = 3/4; c = 0； D=4 .

t= Dx？ e^2x

t′=2dxe^2x+ 2dx？ e^2x

t″=2De^2x+8Dxe^2x+ 4Dx？ e^2x]

t″-4t′+4t

=[2De^2x+8Dxe^2x+ 4Dx？ e^2x]-4[2Dxe^2x+ 2Dx？ e^2x]+4Dx？ e^2x

=2De^2x+(8-8)Dxe^2x+(4-8+4)Dx？ e^2x

=8e^2x，

2D=8，D=4