Variation of Parameters

[JeremyK]

Can someone check my work here? The general solution I came up seems to be a much more complex than all the examples the instructor has give us so I am thinking I may have made a mistake somewhere along the line.

y'' - 4y' + 4y = (x + 1)e^2x
m2 - 4m + 4 = 0
m = 2 m = 2
yc(x) = c1e^(2x) + c2xe^(2x)

yp(x) = V1(x)e^(2x) + V2(x)xe^(2x)
y'p(x) = 2V1(x)e^(2x) + 2V2(x)e^(2x) + V1'(x)e^(2x) + V2'(x)xe^(2x)
V1'(x)e^(2x) + V2'(x)xe^(2x) = 0
yp'(x) = 2V1(x)e^(2x) + 2V2(x)e^(2x)
y''p(x) = 4V1(x)e^(2x) + 4V2(x)e^(2x) + 2V1'(x)e^(2x) + 2V2'(x)e^(2x)

substitute that back in and end up with

2V1'(x)e^(2x) + 2V2'(x)e^(2x) = (x+1)e^(2x)

V1'(x)e^(2x) + V2'(x)xe^(2x) = 0
2V1'(x)e^(2x) + 2V2'(x)e^(2x) = (x + 1)e^(2x)

Do the determinants

V1'(x) =
|--------0----------xe^(2x)----|
|---(x+1)e^(2x)---2e^(2x)---|
---------------------------------------
|-----e^(2x)--------xe^(2x)---|
|-----2e^(2x)--------2e^(2x)--|

= -(x^2 + x)/(2-2x)

and

V2'(x) =
|-----e^(2x)------------0------------|
|-----2e^(2x)-----(x + 1)e^(2x)---|
---------------------------------------------
|-----e^(2x)--------------xe^(2x)---|
|-----2e^(2x)--------------2e^(2x)--|

= (x)/(2 - 2x)

Integrate, V1 = (x^2)/4 + X + ln(x + 1)
V2 = 1/2(-x - ln(2-2x) + 1)

yp(x) = ((x^2)/4 + x + ln(x-1))e^(2x) + (1/2(-x -ln(2-2x)+ 1)xe^(2x))

General Solution

y=c1e^(2x)+c2xe^(2x)+((x^2)/4+x+ln(x-1))e^(2x)+(1/2(-x-ln(2-2x)+1)xe^(2x))


Thanks for looking!

 

[galactus]

I will step through this. I believe your determinants are a little discombobulated. It'll be easier for you to spot the problem.

\(\displaystyle \L\\y''-4y'+4y=(x+1)e^{2x}\)

Do the auxiliary equation:

\(\displaystyle \L\\m^{2}-4m+4=(m-2)^{2}=0\)

We have \(\displaystyle \L\\y_{c}=C_{1}e^{2x}+C_{2}xe^{2x}\).

\(\displaystyle \L\\y_{1}=e^{2x}\) and \(\displaystyle \L\\y_{2}=xe^{2x}\)

Figure the Wronskian:

\(\displaystyle \L\\W(e^{2x},xe^{2x})=\)

\(\displaystyle \L\\e^{2x}\ xe^{2x}\)
\(\displaystyle \L\\2e^{2x}\ 2xe^{2x}+e^{2x}\)

\(\displaystyle \L\\=e^{4x}\)


Now,

\(\displaystyle \L\\W_{1}=\)

\(\displaystyle \L\\\begin{array}{cc}0&xe^{2x}\\(x+1)e^{2x}&2xe^{2x}+e^{2x}\end{array}\)

=\(\displaystyle \L\\-(x+1)xe^{4x}\)

\(\displaystyle \L\\W_{2}=\)

\(\displaystyle \L\\\begin{array}{cc}e^{2x}&0\\2e^{2x}&(x+1)e^{2x}\end{array}\)

\(\displaystyle \L\\=(x+1)e^{4x}\)


\(\displaystyle \L\\u'_{1}=\frac{W_{1}}{W}=\frac{-\(x+1)xe^{4x}}{e^{4x}}
=-x^{2}-x\)

\(\displaystyle \L\\u'_{2}=\frac{W_{2}}{W}=\frac{(x+1)e^{4x}}{e^{4x}}
=x+1\)

So, integrating gives:

\(\displaystyle \L\\u_{1}=\frac{{-}x^{3}}{3}-\frac{x^{2}}{2}\)

\(\displaystyle \L\\u_{2}=\frac{x^{2}}{2}+x\)

So, (we're getting there):

\(\displaystyle \L\\y_{p}=(\frac{-x^{3}}{3}-\frac{x^{2}}{2})e^{2x}+(\frac{x^{2}}{2}+x)xe^{2x}=(\frac{x^{3}}{6}+\frac{x^{2}}{2})e^{2x}\)

So, finally:

\(\displaystyle \L\\y=y_{c}+y_{p}=C_{1}e^{2x}+C_{2}xe^{2x}+(\frac{x^{3}}{6}+\frac{x^{2}}{2})e^{2x}\)

WHEW, please check my work. It's been awhile since I done VOP.