Solving the DE

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  • #1
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Hello!

Here's a question I was doing but I'm not sure if I'm right...

It says "solve" the following differential equation:
[tex](x^2 +1) \frac{dy}{dx} = 0[/tex]

So, this is what I've done:

We put it in form [tex]\frac{dy}{dx} +p(x)y = q(x)[/tex]

Then we devide it by x^2 +1 to obtain:
[tex]\frac{dy}{dx} + \frac{x}{x^2 +1} y = \frac{1}{x^2 +1} \times 0[/tex]

you know , in standard form where p(x) = x/x^2 +1

Integrating factor is:
[tex]exp[\int p(x) dx = exp[\frac{x dx}{x^2 +1}[/tex]
[tex]= exp[ln(x^2 +1)^{2}][/tex]
[tex]= (x^2 +1)^{2}[/tex]

Multiplying the DE by the integrating factor;
[tex](x^2 +1)^{2} \frac{dy}{dx} +x(x^2 +1)y = x . (x^2 +1)^{2}[/tex]

Now we integrate:
[tex]\frac{d}{dx} [(x^2 +1)^{2} y] = x^3 +x[/tex]

=> [tex][(x^2 +1)^{2} y] = \frac{x^4}{4} + \frac{x^2}{2} + c [/tex]

Am I right? Please correct me if I'm wrong...

Thank you,
 

Answers and Replies

  • #2
Vid
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The only answer is y = c.

You have two things being multiplied together equaling zero. This means that either (x^2+1) = 0 or dy/dx = 0. In the first case, the equation is only true when x = +-i which isn't very useful since we solve DEs over an interval and would also allow for y to be any function of x (again not very useful). In the second case, dy/dx = 0 thus y = c is the only solution.

You can see your solution doesn't work by just plugging it into the original equation.
 
  • #3
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Yes, maybe but could you please show me how the actual calculation goes till we get to y = c? :rolleyes:

Thank you.
 
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  • #4
Vid
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dy/dx = 0 from my explanation above. The only (continuous) functions with zero derivatives are constant functions.
 
  • #5
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[tex](x^2 +1) \frac{dy}{dx} = 0[/tex]


The correct answer at the back of the book is [tex]y = \frac{c}{\sqrt{x^2 +1}}[/tex].

I get your explaination, Vid. But the question is asking me to show working and calculations.
 
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  • #6
HallsofIvy
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[tex](x^2 +1) \frac{dy}{dx} = 0[/tex]


The correct answer at the back of the book is [tex]y = \frac{c}{\sqrt{x^2 +1}}[/tex].

I get your explaination, Vid. But the question is asking me to show working and calculations.
No, you haven't understood what he is saying. Vid is trying to tell you is that, since x2+ 1 is never 0, you can divide on both sides by it to get dy/dx= 0 which, integrating, gives y= c, a constant. That should be pretty simple mathematics.
[itex]y(x)= c\sqrt{x^2+1}[/itex] satisfies the equation [itex](x^2+1)dy/dx= xy[/itex], not 0.
 
  • #7
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Yes, Thanks. :smile:

2) Solve: [tex]\frac{dy}{dx} + 2xy = x[/tex]

This is what I think;

[tex]\int 2x dx = x^2[/tex]

So we multiply both sides by [tex]e^{x^2}[/tex]

[tex]\frac{d}{dx} (e^{x^2} y) = xe^{x^2}[/tex]

[tex]e^{x^2}y = \frac{1}{2}e^{x^2} + c[/tex]

=> [tex]y = c e^{x^2} - \frac{1}{2}[/tex]

c is a constant.

I appreciate that if you could tell me if I made any errors or mistakes..
 
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  • #8
Defennder
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[tex]e^{x^2}y = \frac{1}{2}e^{x^2} + c[/tex]
It's correct up to here.

=> [tex]y = c e^{x^2} - \frac{1}{2}[/tex]
This is incorrect. You were dividing the RHS by [tex]e^{x^2}[/tex] right? You should get [tex]\frac{1}{2} + ce^{-x^2}[/tex] instead.
 
  • #9
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[tex]\frac{1}{2} + ce^{-x^2}[/tex] or [tex]\frac{1}{2} + ce^{x^2}[/tex]?

Anyway, Thanks a lot for your attention!
 
  • #10
Defennder
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It's the first one. That is what you get when you divide c by [tex]e^{x^2}[/tex]
 
  • #11
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Of course!
Many thanks...
 
  • #12
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For the first question, if the question was: [tex](x^2 +1) \frac{dy}{dx} +xy = 0[/tex]
then the solution would be: [tex]y = \frac{c}{\sqrt{x^2 +1}}[/tex]

Hmm, so how do we get that? (that was my question but I missed out +xy).

Does any of my working in the very above post make any sense?

I don't know how do this question so I get [tex]y = \frac{c}{\sqrt{x^2 +1}}[/tex] ...
 
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  • #13
Defennder
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What's the integrating factor in this case? Remember you must first express the DE in the form: [tex]\frac{dy}{dx} + P(x)y = Q(x)[/tex]

Your answer appears to be correct, so what's the problem?
 
  • #14
HallsofIvy
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Yes, Thanks. :smile:

2) Solve: [tex]\frac{dy}{dx} + 2xy = x[/tex]
So [tex]\frac{dy}{dx}= x- 2xy= x(1- 2y)[/itex]
[tex]\frac{dy}{1- 2y}= xdx[/itex]

This is what I think;

[tex]\int 2x dx = x^2[/tex]

So we multiply both sides by [tex]e^{x^2}[/tex]

[tex]\frac{d}{dx} (e^{x^2} y) = xe^{x^2}[/tex]

[tex]e^{x^2}y = \frac{1}{2}e^{x^2} + c[/tex]

=> [tex]y = c e^{x^2} - \frac{1}{2}[/tex]

c is a constant.

I appreciate that if you could tell me if I made any errors or mistakes..
 
  • #15
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What's the integrating factor in this case? Remember you must first express the DE in the form: [tex]\frac{dy}{dx} + P(x)y = Q(x)[/tex]

Your answer appears to be correct, so what's the problem?
Yes I did put it in the standard form and I found the integrating factor

[tex](x^2 +1) \frac{dy}{dx} +xy = 0[/tex]

standard form : [tex]\frac{dy}{dx} +p(x)y = q(x)[/tex]

Then we devide it by x^2 +1 to obtain:
[tex]\frac{dy}{dx} + \frac{x}{x^2 +1} y = \frac{1}{x^2 +1} \times 0[/tex]

Integrating factor is:
[tex]exp[\int p(x) dx = exp[\frac{x dx}{x^2 +1}[/tex]

[tex]= exp[ln(x^2 +1)^{2}][/tex]
[tex]= (x^2 +1)^{2}[/tex]

multiplying it by the integrating factor;
[tex](x^2 +1)^{2} \frac{dy}{dx} +x(x^2 +1)y = x . (x^2 +1)^{2}[/tex]

Now we integrate: [tex]\frac{d}{dx} [(x^2 +1)^{2} y] = x^3 +x[/tex]


I get: [tex][(x^2 +1)^{2} y] = \frac{x^4}{4} + \frac{x^2}{2} + c [/tex]

But this is wrong since the correct answer should be: [tex]y = \frac{c}{\sqrt{x^2 +1}}[/tex]

I don't know how to get that.:cry:
 
  • #16
Vid
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Int(x/(x^2+1) = 1/2ln(x^2+1) = ln|(x^2+1)^(1/2)| not squared. The right side is just zero so the integral is just c; solve for y to get the right answer. Also this DE is easier as a separable equation instead of a first order linear.
 
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  • #17
HallsofIvy
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There is no need to use integrating factors: you have a separable equation.
[tex](x^2 +1) \frac{dy}{dx} +xy = 0[/tex]
Subtractiong xy from both sides and dividing on both sides by [itex]x^2+ 1[/itex]
[tex]frac{dy}{dx}= \frac{xy}{x^2+ 1}= \right(\frac{x}{x^2+1}\right) y[/tex]
[tex]ydy= \frac{x dx}{x^2+1}[/tex]
 

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