# variation of parameters

• May 25th 2008, 07:28 PM
taurus
variation of parameters
I am a bit confused on variation of parameters for solving differential equations. I understand the methods of coefficients but variation of parameters is a bit confusing. I would i use it to solve this:

y'' - 5y' = -2x + 2

Like how would i find independent solutions u1(x), u2(x) which would help me find derivatives v1¢(x) and v2¢(x) which then i get a general solution. I am confused on the steps and how to solve.

Thanks
Phil
• May 25th 2008, 07:31 PM
mr fantastic
Quote:

Originally Posted by taurus
I am a bit confused on variation of parameters for solving differential equations. I understand the methods of coefficients but variation of parameters is a bit confusing. I would i use it to solve this:

y'' - 5y' = -2x + 2

Like how would i find independent solutions u1(x), u2(x) which would help me find derivatives v1¢(x) and v2¢(x) which then i get a general solution. I am confused on the steps and how to solve.

Thanks
Phil

Have you been taught this technique or are you trying to self-learn it.

The solutions u1 and u2 come from solving y'' - 5y' = 0.
• May 25th 2008, 07:36 PM
taurus
Briefly taught as we mainly focused on method by coefficients. But then i saw a questions like that so i thought i could get some advice on how to use the technique of variations by parameters to solve such questions I posted above.
• May 25th 2008, 07:42 PM
TheEmptySet
Quote:

Originally Posted by taurus
I am a bit confused on variation of parameters for solving differential equations. I understand the methods of coefficients but variation of parameters is a bit confusing. I would i use it to solve this:

y'' - 5y' = -2x + 2

Like how would i find independent solutions u1(x), u2(x) which would help me find derivatives v1¢(x) and v2¢(x) which then i get a general solution. I am confused on the steps and how to solve.

Thanks
Phil

First we need to solve the homogeneous system

$\displaystyle y''-5y=0$

The auxillary equation is

$\displaystyle m^2-5m=0 \iff m(m-5)=0, m=0,m=5$

$\displaystyle y_c(x)=c_1e^{0x}+c_2e^{5x}=c_1+c_2e^{5x}$

Now we can set up the wronskian

$\displaystyle W=\begin{vmatrix}1 && e^{5x} \\ 0 && 5e^{5x} \end{vmatrix}=5e^{5x}$

$\displaystyle W_1=\begin{vmatrix}0 && e^{5x} \\ -2x+2 && 5e^{5x} \end{vmatrix}=(2x-2)e^{5x}$

$\displaystyle W_2=\begin{vmatrix}1 && e^{5x} \\ 0 && -2x+2 \end{vmatrix}=-2x+2$

So finally

$\displaystyle v_1'(x)=\frac{W_1}{W}$

$\displaystyle v_2'(x)=\frac{W_2}{W}$

Just integrate from here.

I will leave the rest to you.

Good luck. $\displaystyle \emptyset$
• May 25th 2008, 07:50 PM
taurus
What exactly did you do with the wronskian? Iv never heard of that?
why W, W1, W2?
• May 25th 2008, 08:03 PM
TheEmptySet
Quote:

Originally Posted by taurus
What exactly did you do with the wronskian? Iv never heard of that?
why W, W1, W2?

for an equation

$\displaystyle y''+a(x)y'+b(x)y=f(x)$

When I found the complimentry solution

$\displaystyle y_c=c_1+c_2e^{5x}$

This is

$\displaystyle y_c=c_1u_1(x)+c_2u_2(x)$

The wronskian is

$\displaystyle W=\begin{vmatrix}u_1(x) && u_2(x) \\ u_1'(x) && u_2'(x) \end{vmatrix}$

$\displaystyle W_1=\begin{vmatrix}0 && u_2(x) \\ f(x) && u_2'(x) \end{vmatrix}$

$\displaystyle W_2=\begin{vmatrix}u_1(x) && 0 \\ u_1'(x) && f(x) \end{vmatrix}$

Now the particular solution to the equation is

$\displaystyle y_p=u_1\cdot \int\frac{W_1}{W}dx+u_2 \cdot \int\frac{W_2}{W}dx$

I hope this clears it up.

Good luck.
• May 26th 2008, 12:15 AM
taurus
For v1' and v2' above is it:
v1 = ((2x-2)e^(5x))/(5e^(5x)) integrated
which gives
v1' = ((1/5)*(x-2)*x)

v2 = (-2x+2)/(5e^(5x)) integrated
gives
v2' = ((2/125)*e^(-5*x) * (5*x-4))

Is that correct? or am I differentiating W1/W and W2/w?

thanks
• May 26th 2008, 03:16 AM
galactus
You can check this out. I outlined a VP problem in post #6.

http://www.mathhelpforum.com/math-he...eous-help.html
• May 26th 2008, 06:59 PM
taurus
Quote:

Originally Posted by TheEmptySet

$\displaystyle v_1'(x)=\frac{W_1}{W}$

$\displaystyle v_2'(x)=\frac{W_2}{W}$

$\displaystyle \emptyset$

That is what i did and doesnt seem to be right. I integrated W1/W and W2/W

For v1' and v2' above is it:
v1' = ((2x-2)e^(5x))/(5e^(5x)) integrated
which gives
v1 = ((1/5)*(x-2)*x)

v2' = (-2x+2)/(5e^(5x)) integrated
gives
v2 = ((2/125)*e^(-5*x) * (5*x-4))
What am I doing wrong?
• May 26th 2008, 07:51 PM
TheEmptySet
$\displaystyle v_1=\int\frac{1}{5} \left( 2x-2 \right)dx=\frac{1}{5}x^2-\frac{2}{5}x$

$\displaystyle v_2=\int -\frac{2}{5}(x+1)e^{-5x}dx=\left[\frac{2}{25}x+\frac{12}{125} \right]e^{-5x}$

So the particular solution is

$\displaystyle y_p=u_1v_1+u_2v_2=1\left[\frac{1}{5}x^2-\frac{2}{5}x\right]+e^{5x}\left[\frac{2}{25}x+\frac{12}{125} \right]e^{-5x}=\frac{1}{5}x^2-\frac{8}{25}x+\frac{12}{125}$

Now when you combine this with the complimentry solution we get

$\displaystyle y=y_c+y_p=c_1+c_2e^{5x}+\frac{1}{5}x^2-\frac{8}{25}x+\frac{12}{125}$

collecting and relabeling the constants we get the solution

$\displaystyle y=c_3+c_2e^{5x}+\frac{1}{5}x^2-\frac{8}{25}x$
• May 26th 2008, 08:02 PM
taurus
I dont think your V1 and V2 is correct because i entered it into computer and it said its wrong?
• May 26th 2008, 08:14 PM
TheEmptySet
Quote:

Originally Posted by taurus
I dont think your V1 and V2 is correct because i entered it into computer and it said its wrong?

I disagree with your computer. I did the first computations by hand first and now I have checked them in Maple. Maple has written them differently, but they are equal to what I posted before.

Here is a screen shot. What software are you using?

Attachment 6520
• May 26th 2008, 08:19 PM
taurus
hmmmm yea Im also using maple.

The questions am trying to answer this:
Now assume that the solution of the original d.e. is of form

y(x) = u1v1 + u2+v2

where u1, u2 are the functions you entered in the first part. Determine the derivatives v1'(x) and v2'(x).
• May 26th 2008, 08:28 PM
TheEmptySet
Quote:

Originally Posted by taurus
hmmmm yea Im also using maple.

The questions am trying to answer this:
Now assume that the solution of the original d.e. is of form

y(x) = u1v1 + u2+v2

where u1, u2 are the functions you entered in the first part. Determine the derivatives v1'(x) and v2'(x).

As I said in post number four

$\displaystyle v_1'=\frac{W_1}{W}$
and
$\displaystyle v_2'=\frac{W_2}{W}$

Am I misunderstanding your question. Are you asking for a proof of why the method works?
• May 26th 2008, 09:52 PM
taurus
hmmm yea which is what i tried but computer says it wrong, hmm i dont know anymore