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I need the steps

  • Thread starter madahmad1
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  • #1
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Homework Statement


Problem statement: Find the solution to the differential equation.

1) dy/dx= x^3-2y/x solution is: y= c/x^2 + x^3/5 the equation is linear but i do not know the steps.

The same goes for the following:

2. (x+y)dx - (x-y)dy=0 Solution is: arctan(y/x) - ln (square root of)x^2+y^2=c

6. xdy/dx +xy=1-y y(1)=0 solution is: y= x^-1(1-e^1-x)

8. xdy/dx +2y= sinx/x y(2)=1 solution is: (4+cos2-cosx)/x^2

12. dy/dx +y = 1/1+e^x solution is: y= ce^-x + e^-xln(1+e^x)

15. (e^x +1)dy/dx= y-ye^x solution is: y= c/cosh^2(x/2)

17. dy/dx= e^2x + 3y solution is: y= ce^3x-e^2x

20. y`= e^x+y solution is: e^x + e^-y=c

22. dy/dx= x^2-1/y^2+1 y(-1)=1 solution is: y^3+3y-x^3+3x=2

30. dy/dx= y^3/1-2xy^2 y(0)=1 solution is: xy^2 -ln[y]=0

31. (x^2y+xy-y)dx + (x^2y-2x^2)dy=0 solution is:[x+ln[x]+x^-1+y-2ln[y]=c


Please show me how you solve these problems. I know that 15,20,22,&31 are separable. The rest are linear except for 2, its homogeneous.






The Attempt at a Solution



For linear equations I am to find the integrating factor which is e to the power of the coefficient of y mulitplied by t. (e^xt) Then I multiply this factor with the equation and integrate to find the value of y. For seperable I am supposed to split the equation with x on one side and y on the other then integrate to find y. I tried doing that but could not figure out how to get the solutions that my professors gave me. Any help please?
 

Answers and Replies

  • #2
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any help please?
 
  • #3
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Pick one or two problems and show us what you have tried, and we'll take it from there. I can't speak for everyone on this forum, but I'm not particularly excited about doing 10 or 11 problems for you.
 
  • #4
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Ok, for problem 1, I re-arranged the equation so that dy/dx -2y = x\2 then I found that the integrating factor is e^-2t since -2 is the coefficient of y. Then I integrate that but i do not know how the Dr. came up with the solution that he got because I got something different.

This is the correct solution: y= c/x^2 + x^3/5


For number 22, I got (y^2 +1)dy= (x^2-1)dx
then, (y^3/3) +y= (x^3/3) -x
then (y^3/3) +y -(x^3/3)+x
then multiply all that by 3

y^3 +3y -x^3+3x=0 boundary conditions are y(-1)= 1

so I replaced y with -1 in the equation and x with 1 and that way I would get c. But c did not equal 2, why is that? Please help
 
  • #5
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For 17 integrating factor is e^-3t so I re-arrange the equation and multiply it by e^-3t. I got
dy/dx e^-3t - 3ye^-3t= e^2x-3t

Then I integrate it, but how do you get y= ce^3x-e^2x by integration?
 
  • #6
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Ok, for problem 1, I re-arranged the equation so that dy/dx -2y = x\2 then I found that the integrating factor is e^-2t since -2 is the coefficient of y.
How did you get dy/dx - 2y = x/2 starting from dy/dx = x3 - 2y/x?

If you rearrange the equation to get all the y and dy/dx terms on one side, you get this:
dy/dx + 2y/x = x3

If I'm not mistaken, the integrating factor is 2ln(x). If you multiply both sides of the equation by this integrating factor, what you have on the left is the derivative of y*2ln(x), so when you integrate it, you get y*2ln(x).
 
  • #7
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For 17 integrating factor is e^-3t so I re-arrange the equation and multiply it by e^-3t. I got
dy/dx e^-3t - 3ye^-3t= e^2x-3t

Then I integrate it, but how do you get y= ce^3x-e^2x by integration?
You have too many variables, with y, x, and t. You should have just two variables, y and x, since the original problem is given in terms of y and x. With that change, your integrating factor is e-3x.

Then, the left side should look like dy/dx * e-3x - 3y e-3x, which happens to be the derivative with respect to x of ye-3x. If you integrate the derivative of ye-3x, you get ye-3x (plus a constant, but you can take care of that on the other side of the equation).

So, what you have is ye-3x = [itex]\int e^{2x}e^{-3x} dx[/itex]. Simplify the integrand on the right before integrating. After you have done the integration, solve for y. That's the solution you're trying to find.
 
  • #8
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For problem 1 I divided by x so x^3 divided by x was x^2 but I`m not sure if that is correct. The final answer should be y= c/x^2 + x^3/5 but I do not know how that was given. Can you show me?
 
  • #9
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For problem 1 I divided by x so x^3 divided by x was x^2 but I`m not sure if that is correct. The final answer should be y= c/x^2 + x^3/5 but I do not know how that was given. Can you show me?
I am sure that that is NOT correct. The equation you started with is dy/dx= x^3-2y/x. Rearranging this gives dy/dx + 2y/x = x^3. If you divide both sides of the equation by x, you do get x^2 on the right, but you get a mess on the left side that is not help.

Don't worry so much about what the final answer is: worry instead about understanding how to solve these problems. As I said in post 6, I think the integrating factor is 2ln(x), ln(x^2) (they are the same, if x > 0).

Multiply both sides of your initial equation by the integrating factor. Read what I wrote in post 7. The technique in problem 1 and problem 17 is the same.
 
  • #10
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yes I know both are linear equations. But, can someone please give me a clear step-by-step on how to do it? I showed you what my method and answer was but couldn`t get it to be the same as the correct answer.
 
  • #11
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4,838
What part of these don't you understand?
For problem 17:
You should have just two variables, y and x, since the original problem is given in terms of y and x. With that change, your integrating factor is e-3x.

Then, the left side should look like dy/dx * e-3x- 3ye-3x, which happens to be the derivative with respect to x of ye-3x. If you integrate the derivative of ye-3x, you get ye-3x(plus a constant, but you can take care of that on the other side of the equation).

So, what you have is ye-3x = [itex]\int e^{2x}e^{-3x} dx[/itex]
. Simplify the integrand on the right before integrating. After you have done the integration, solve for y. That's the solution you're trying to find.
For problem 1:
As I said in post 6, I think the integrating factor is 2ln(x), ln(x^2) (they are the same, if x > 0).

Multiply both sides of your initial equation by the integrating factor. Read what I wrote in post 7. The technique in problem 1 and problem 17 is the same.
The title of the thread you started is "I need the steps." I am giving you a broad outline of how to get the solutions to a couple of these problems. Now it's up to you to take the steps I've outlined.
 
  • #12
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For problem 1. I re-arranged and got y`(x) +2y = x^3. Thus, the integrating factor is e^2x, is that correct? Now, I multiply by e^2x and continue, correct? Or is this method wrong?
 
  • #13
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4,838
For problem 1. I re-arranged and got y`(x) +2y = x^3. Thus, the integrating factor is e^2x, is that correct? Now, I multiply by e^2x and continue, correct? Or is this method wrong?
Here's what you started with:
dy/dx= x^3-2y/x

How do you get dy/dx + 2y = x^3 from that? If you add 2y/x to both sides, you get dy/dx + 2y/x = x^3

If you then multiply by x on both sides, you get
xdy/dx + 2y = x^4, not dy/dx + 2y = x^3.
 
  • #14
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you multiply both sides by x and then move 2y to the other side.
 
  • #15
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And then you get xdy/dx + 2y = x^4, not dy/dx + 2y = x^3, as you show. I have told you what the integrating factor for problem 1 is in posts 6, 9, and 11. The difficulties you are having finding the integrating factors seems to be from a weakness in some pretty basic algebra, so unless I'm wrong, you're going to have a real tough time completing one of these problems, let alone all of them.
 
  • #16
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sorry I missed that, now I multiply all that by e^2. After that, I integrate again right?
 
  • #17
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sorry I missed that, now I multiply all that by e^2. After that, I integrate again right?
NO, NO, NO! The integrating factor is NOT e2x!
 
  • #18
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why not? Is it e^integral 2x or e^x^2/2 ?
 
  • #19
HallsofIvy
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why not? Is it e^integral 2x or e^x^2/2 ?
Yes, it is e^integral 2x. That is not e^(x^2/2).
 
  • #20
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Because you don't have a differential equation of the form y' + p(x)y = q(x), that's why not.

What you have is xy' + 2y = x^4.

What do you need to do to get this equation to look like y' + p(x)y = q(x)?
 
  • #21
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divide by x to get rid of the xy`. Si it becomes y` + 2y/x= x^3
 
  • #22
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Yes, it is e^integral 2x. That is not e^(x^2/2).
This is incorrect, and the integrating factor I have stated in several posts in this thread is incorrect, also. The integrating factor is [tex]e^{\int \frac{2dx}{x}} = e^{2 ln(x)} = e^{ln(x^2)} = x^2.[/tex]
 
  • #23
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4,838
divide by x to get rid of the xy`. Si it becomes y` + 2y/x= x^3
Yes. Now multiply both sides by the integrating factor (x^2). On the left side you have
y' x2 + 2y/x * x2, and on the right side you have x5

The left side should be recognizable as the derivative of some product. In other words,
y' x2 + 2y/x * x2 = d/dx(??).

If you integrate this derivative, the left side will be whatever is represented by ??. Integrate the right side as well, and then solve algebraically for y.
 
  • #24
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is that product 2x?
 
  • #25
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Ok, for problem 6, I got dy/dx + 2y/x = 1/x by re-arranging and dividing by x. Since 2y/x is the coefficient then the integrating factor is the same as before x^2.

multiply all sides by x^2 I get, dy/dx(x^2)+ 2xy = x, next we integrate so

x^3/3+c = x^2/2 but how do I get the answer I provided in the beginning with this result?
 

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