Linear Differential Equation with a substitution.

In summary: Thanks in advance.Differentiating y(x) = sinx + 1/u(x)must this be done implicitly? I get confused as u(x) is a function and not a variable.No, you just need to use the chain rule.
  • #1
Rake-MC
325
0
Wow I am so very bad at differential equations.. :(

The problem

Here is the exact problem I'm given: http://img229.imageshack.us/img229/9025/10640260bs5.jpg [Broken]

attempt at a solution

I'm guessing that I need to differentiate y(x) that I am given and substitute that into the left hand side and then put the y function [sinx + 1/u(x)] as y in the right hand side (y^2).
Then re-arrange for u(x) and differentiate to get du/dx and hopefully it will be the third given equation.

After that I think I just have to solve that ODE and get it back in terms of y.

Assuming that is what I have to do, sadly I cannot even get that far..

differentiating y(x) = sinx + 1/u(x)
must this be done implicitly? I get confused as u(x) is a function and not a variable.

I can see that the third equation must be solved linearly such that dy/dx + utan(x) = -1/2 sec(x)

And I (think) that I solved the integrating factor to be 1/cos(x).

But I am so bad at differential equations and the sources on the internet seem to be so confusing that I'm quite stuck from here but this is what I have:

(1/cos(x))*du/dx + u*(1/cos(x))*tan(x) = -(1/2)*(1/cos(x))*sec(x)=d/dx(u/cos(x)) <- multiplying by integrating factor and copying form of examples from the net

u/cos(x) = integral of -(1/2)*(1/cos(x))*sec(x)

and I know I'm way off here so I think I might as well just stop...

Also it would be great if someone could point me in the direction of a site that shows how to type all those math symbols such as the integral sign.

Thanks in advance. A lot.
 
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  • #2
Rake-MC said:
I'm guessing that I need to differentiate y(x) that I am given and substitute that into the left hand side and then put the y function [sinx + 1/u(x)] as y in the right hand side (y^2).
Then re-arrange for u(x) and differentiate to get du/dx and hopefully it will be the third given equation.
Good guess :smile:

Rake-MC said:
differentiating y(x) = sinx + 1/u(x)
must this be done implicitly? I get confused as u(x) is a function and not a variable.

I don't know what you call "implicitly" but I don't think so. All you need is the chain rule:
[tex]\frac{d}{dx} \frac{1}{u(x)} = \frac{d (1/u)}{du} \frac{du}{dx}[/tex]
 
  • #3
Thanks CompuChip.

For dy/dx I got cos(x) - 1/u^2

so that's on the left side, for the right side I subbed in sin(x) + 1/u

but it just gets really messy, and I can't get u on its own.. This is probably one of the most simple things to do but I just can't do it.

Maybe I should drop out of engineering...
 
  • #4
Rake-MC said:
For dy/dx I got cos(x) - 1/u^2

[...]

but it just gets really messy, and I can't get u on its own..

You still forgot the chain rule:

dy/dx = cos(x) - u' / u^2,
where u' denotes differentation of u w.r.t. x.
And you don't want u on its own, but du/dx = u'.
 
  • #5
Thanks tons!
I can't show how grateful I am for being so patient with me. I am so slow with this kind of thing..

Now I'm 99% sure I have to do a linear differential equation to get the solution, but doing that will probably result in colossal failure on my part.
As I said earlier I'm sure the integrating factor is 1/cos(x), and I multiplied it through my differential equation such that I had it in the form of I(x)dy/dx + I(x)P(x) = Q(x)
but I don't really know what to do from here.
Sites that I've looked at seem to use terminology and symbols that are far beyond me..
 
  • #6
Okay, linear differential equations confuse the hell out of me, but by any chance is the answer:

[tex]u = -\frac{1}{2} \int sec(x)sec(x)dx = -\frac{1}{2} tan(x) + C [/tex]
[tex] y(x) = sin(x) + \frac{1}{u(x)} [/tex]
[tex]y(x) = sin(x) - \frac{2}{tan(x)+C}[/tex]

Did I do this right?
Cheers
 
  • #7
Somehow I don't think that this is right, because I have dy/dx, subbed in u, took the derivative of that. Did the differential equation, but shouldn't I need to do another differential equation to reach y(x)? It confuses me because I'm given dy/dx as well as y(x) in terms of u(x).
Thanks
 
  • #8
Okay, I think that I have done something wrong in that it's not:
[tex]
u = -\frac{1}{2} \int sec(x)sec(x)dx
[/tex]

but rather [tex]
I(x)u = -\frac{1}{2} \int sec(x)sec(x)dx = -\frac{1}{2} tan(x) + C
[/tex]

where [tex] I(x) = sec(x) [/tex]
so

[tex]
u = -\frac{1}{2} tan(x)cos(x) + C
[/tex]

Then [tex]
y(x) = sin(x) - \frac{2}{tan(x)cos(x)+C} = sin(x) - 2cot(x)sec(x)
[/tex]

If anyone could confirm this, it would be most greatly appreciated.
Cheers.
 
  • #9
Bump thx.
 
  • #10
Yeap, you've got it right. Just one thing, where did your constant go for the final answer? Btw, we are on the same course doing the same subject. I found the assignment really difficult esp question 3.
 
  • #11
Cheers jeffreylze,
well my constant of integration is [tex] \frac{1}{cos(x)} [/tex]

which is the [tex] sec(x) [/tex]

in the final answer:

[tex] sin(x) - 2cot(x)sec(x) [/tex]

And yes a very difficult assignment, luckily we got over the break for it!
 

1. What is a linear differential equation with a substitution?

A linear differential equation with a substitution is a type of differential equation where a substitution is made to transform the equation into a linear form. This allows for easier solving of the equation using methods such as separation of variables or integrating factors.

2. How do you solve a linear differential equation with a substitution?

To solve a linear differential equation with a substitution, you first identify the substitution that will transform the equation into a linear form. Then, apply the substitution to the original equation and solve for the new variable. Finally, substitute back to find the solution to the original equation.

3. What are the advantages of using a substitution to solve a linear differential equation?

Using a substitution to solve a linear differential equation can simplify the equation and make it easier to solve. It can also help to reveal patterns and relationships within the equation that may not have been apparent before the substitution was applied.

4. Are there certain types of linear differential equations that are more suitable for a substitution?

Yes, linear differential equations that have variables that can be easily isolated and substituted are more suitable for a substitution. For example, equations with separable variables or with terms that can be factored.

5. Can a linear differential equation with a substitution have multiple solutions?

Yes, a linear differential equation with a substitution can have multiple solutions. This is because the substitution can sometimes introduce additional constants or variables that can lead to different solutions. It is important to check the solutions obtained using the substitution to ensure they are valid for the original equation.

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