# Math Help - variation of parameters

1. ## 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

2. 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.

3. 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.

4. 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

$y''-5y=0$

The auxillary equation is

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

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

Now we can set up the wronskian

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

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

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

So finally

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

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

Just integrate from here.

I will leave the rest to you.

Good luck. $\emptyset$

5. What exactly did you do with the wronskian? Iv never heard of that?
why W, W1, W2?

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

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

When I found the complimentry solution

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

This is

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

The wronskian is

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

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

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

Now the particular solution to the equation is

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

I hope this clears it up.

Good luck.

7. 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

8. You can check this out. I outlined a VP problem in post #6.

http://www.mathhelpforum.com/math-he...eous-help.html

9. Originally Posted by TheEmptySet

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

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

$\emptyset$
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?

10. $v_1=\int\frac{1}{5} \left( 2x-2 \right)dx=\frac{1}{5}x^2-\frac{2}{5}x$

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

So the particular solution is

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

Now when you combine this with the complimentry solution we get

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

collecting and relabeling the constants we get the solution

$y=c_3+c_2e^{5x}+\frac{1}{5}x^2-\frac{8}{25}x$

11. I dont think your V1 and V2 is correct because i entered it into computer and it said its wrong?

12. 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?

13. 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).

14. 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

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

Am I misunderstanding your question. Are you asking for a proof of why the method works?

15. hmmm yea which is what i tried but computer says it wrong, hmm i dont know anymore