Final step of maths question please help

Hi, currently doing a second differential equation:

y''-4y'+4y=e^2x

ive found the auxiliary equation and its solutions:
m^2-4m+4=0
therefore m must = 2

giving me the complementary function (A+Bx)e^2x

Im struggling to find the particular integral to finish this off - ive tried Yp(X)=ae^2x but the terms just cancel, the same with Yp(X)=axe^2x

any help finding this particular integral would be greatly appreciated

(edited 7 years ago)

Reply 1

Pangol

15

If you expand your CF, you get Ae^2x + Bxe^2x. So, your CF has a term in e^2x AND a term in xe^2x. Your trial function therefore has to be of the form C(x^2)(e^2x).

Original post by ComplexNerd

Hi, currently doing a second differential equation:

y''-4y'+4y=e^2x

ive found the auxiliary equation and its solutions:
m^2-4m+4=0
therefore m must = 2

giving me the complementary function (A+Bx)e^2x

Im struggling to find the particular integral to finish this off - ive tried Yp(X)=ae^2x but the terms just cancel, the same with Yp(X)=axe^2x

any help finding this particular integral would be greatly appreciated

Try

y_p(x)=ax^2e^{2x}

Reply 3

ComplexNerd

OP

Original post by Pangol

If you expand your CF, you get Ae^2x + Bxe^2x. So, your CF has a term in e^2x AND a term in xe^2x. Your trial function therefore has to be of the form C(x^2)(e^2x).