Introduction to Solving Differential Equations: Finding a New Variable

In summary, the student is trying to find y' and is having trouble because y-x is squared and not just x or y. They suggest substituting x-y=z and inserting it into the equation, which creates a new equation with y'=z'+1. When solving this equation, they lose z. They try the ln calculation rule and find that A=exp(2c). Finally, they find y=\frac{-1+Ae^{2x}}{1-Ae^{2x}}+x and solve for x.
  • #1
Lindsayyyy
219
0
Hi


Homework Statement



[tex] y'=(y-x)^{2}[/tex]


Homework Equations





The Attempt at a Solution



I have nearly no idea in order to solve this solution. One hint is given: We shall find a new variable which makes it possible to separate the differential equation afterwards.
I'm not sure what's meant by "new variable". Shall I just find something cmpletly new, like let's call it z, or an addition like +x or *x etc.(just for example, I know this specific example doesn't work)?

Thank you for your help
 
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  • #2
First thing I can suggest is posting this in the right place. Differential equations are definitely NOT precalculus
 
  • #3
Lindsayyyy said:
Hi

Homework Statement



[tex] y'=(y-x)^{2}[/tex]

Have you studied homogeneous functions? Your right side is homogeneous of degree 2 so try changing the dependent variable from y to u by y = ux.

[Edit:] Never mind this suggestion. When the equation is put in the form Mdx + Ndy = 0 not both M and N are homogeneous.
 
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  • #4
Since it is the fact that it is y- x that is squared and not just x or y that is the problem, let z= y- x. Then z'= y'- 1 so that y'= z'+ 1= z^2. z'= z^2- 1.
 
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  • #5
When you get the answer, make sure you check it by taking the derivative - it's really cool how it works out! :)
 
  • #6
do the subst:
[tex]
z(x) = y(x) - x
[/tex]
Then the variables in the new equation can be separated.
 
  • #7
I'm still unsure.

I tried the following, I took the substition from HallsofIvy

[tex] y'=z^{2}-1[/tex]

Shall I just integrate both sides now? I think this doesn't work, as I still have y and x after the integration on one side.

And I wonder about the following:

I have [tex]y'=z^{2}-1 [/tex] but when I say I substitute [tex] x-y=z[/tex] and insert it in my given equation I have [tex] y'=z^{2}[/tex] :mad:
 
  • #8
I think you misunderstood. The equation HallsofIvy got to is

[tex]z' = z^2 - 1[/tex]

Not y'.
 
  • #9
You performed the substitution incorrectly. If z = y - x, take d/dx of both sides: dz/dx = dy/dx - 1. Since dy/dx = (y - x)^2 = z^2, we have dz/dx = z^2 - 1.

Now, separate variables and use partial fractions. To separate variables, integrate both sides of f(z) dz = g(x) dx. :)
 
  • #10
Thank you, just to make sure I didn't do any mistakes thus far. I have to do the partial fraction with the following:

[tex] x= \int \frac1 {z^{2}-1} dz[/tex]
 
  • #11
Yes. For simplicity's sake, I'd put your constant of integration (+ C) on the x side.
 
  • #12
Wow, it's really embarrassing right now.

My partial fraction looks like:

[tex] \frac 1 {z^{2}-1} = \frac A {z-1} + \frac B {z+1} [/tex]
[tex]
A= \frac 1 2[/tex]
[tex]
B= -\frac1 2

[/tex]

Now I tried to integrate

[tex]

x=\frac1 2 \int{\frac 1 {z-1}}dz - \frac 1 2 \int{\frac 1 {z+1}}dz [/tex]

getting:

[tex] 2x+2c = ln(z-1)-ln(z+1)[/tex]

but the z gets lost when I *exp on both sides. Where's my mistake. It's so frustrating :cry:
 
  • #13
Don't fret. Recall that ln(a) + ln(b) = ln(ab) and ln(a) - ln(b) = ln(a/b).

Also, when you take the exponential of 2x + 2C, let A = e^(2C) for simplicity. Your left side should be Ae^(2x).
 
  • #14
Thanks for the help and patience :biggrin:

my solution looks like this:

I used the ln calculation rule:

whereas, as you said, my A=exp(2c)

[tex] z=\frac {1-A-e^{2x}} {-1+A+e^{2x}} [/tex]

the only step I have to do now is to resubstitute z with [tex] z= x-y [/tex]

and solve the equation, bringing y on one side, which would look like this:

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

is this correct (and the final answer) ?
 
  • #15
No, but close. For any base k > 0, k^(a + b) does not equal k^a + k^b; it equals (k^a)(k^b).

Also, we set z = y - x, not z = x - y.
 
  • #16
I should really try to concentrate some more :smile:

Another attempt, A=exp(2c) like before

final equation looks like:

[tex] y =\frac{-1-Ae^{2x}} {-1+Ae^{2x}}+x[/tex]
 
  • #17
Yes! Well done. One more thing: you could decimate the abundance of minus signs (*shudder*) by multiplying with -1/-1 to get
[tex]y = \frac{1 + Ae^{2x}}{1 - Ae^{2x}} + x[/tex]
as your final answer.

As an exercise, you should find y' and see that it does indeed equal (y - x)2. See post #5 :)
 
  • #18
Thank you very much :smile:
I'll try that, I tried it for the specific A=0 because that worked quite fast, but I'll do it generally and see how it works. I really need some practice with differential equations.
 

FAQ: Introduction to Solving Differential Equations: Finding a New Variable

1. What is a differential equation?

A differential equation is a mathematical equation that describes the relationship between a function and its derivatives. It is used to model many real-world phenomena, such as motion, growth, and decay.

2. Why do we need to find a new variable to solve a differential equation?

In some cases, the original variable in a differential equation may be difficult to work with or may not lead to a solution. Finding a new variable can simplify the equation and make it easier to solve.

3. How do you find a new variable for solving a differential equation?

There are various methods for finding a new variable, such as substitution, separation of variables, or using an integrating factor. The specific method used depends on the type of differential equation and its characteristics.

4. Can any differential equation be solved by finding a new variable?

No, not all differential equations can be solved by finding a new variable. Some equations may require more advanced techniques or may not have a closed-form solution. It is important to understand the limitations and assumptions of the methods used.

5. How is solving differential equations useful in science?

Solving differential equations allows us to model and understand complex systems in various fields of science, such as physics, biology, economics, and engineering. It helps us make predictions and analyze the behavior of these systems, which can aid in making informed decisions and advancements in our understanding of the world.

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