Solve this Differential Equation: (1+x^2)u"(x) + 2xu'(x) - 2u(x) = 0, u(x) = (c_1)x + (c_2)(1 + x*tan^{-1}(x))
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A series solution works.
Let [imath]u(x)=a_{0}+a_{1}x + a_{2}x^{2}+a_{3}x^{3}+\dots[/imath]
Substitute into the equation and equate coefficients to zero.
You should find that [imath]a_{k+2}=-a_{k}(k-1)/(k+1),[/imath],
that [imath]a_{1}[/imath] is indeterminate, and that [imath]a_{3}=0.[/imath]
The infinite series gets you the x.arctan(x) term.
Let [imath]u(x)=a_{0}+a_{1}x + a_{2}x^{2}+a_{3}x^{3}+\dots[/imath]
Substitute into the equation and equate coefficients to zero.
You should find that [imath]a_{k+2}=-a_{k}(k-1)/(k+1),[/imath],
that [imath]a_{1}[/imath] is indeterminate, and that [imath]a_{3}=0.[/imath]
The infinite series gets you the x.arctan(x) term.
Thank you Bob.
You know what Bob, you are the greatest new member I have ever encountered in this forum. You are suggesting that I may use the power series [imath]\displaystyle \sum_{n=0}^{\infty}a_nx^n[/imath] to solve the problem.
Well, if you know me enough, you would realize that I was the only student in this forum since its creation who has solved differential equations using power series. I did not try this method in this problem because I thought that there might be a faster approach.
One of my remarkable solutions was to the Airy Equation. I have solved it from scratch using power series. Thanks for the try Bob. If I run out of ideas, and did not find a quicker attack, I would try to use your suggestion and solve this problem here in this forum following what you have provided.
An alternative method, but I don't think that it's any 'faster' is 'a reduction of order', [imath]u=x[/imath] is clearly a solution of the equation, so as a second solution try [imath]u=x\phi(x).[/imath]
That will get you a first order equation for [imath]\phi '(x)[/imath]which is solvable, (integrating factor), and that can then be integrated to get [imath]\phi (x).[/imath]
That will get you a first order equation for [imath]\phi '(x)[/imath]which is solvable, (integrating factor), and that can then be integrated to get [imath]\phi (x).[/imath]
Let us check if [imath]u(x) = x[/imath] is a solution.
[imath]u'(x) = 1[/imath]
[imath]u''(x) = 0[/imath]
Plugging in original equation.
[imath](1 + x^2)\times 0 + 2x\times 1 - 2\times x = 0[/imath]
[imath]2x - 2x = 0[/imath]
Confirmed.
Now let us check if [imath]u(x) = x\phi(x)[/imath] is a solution.
[imath]u'(x) = \phi(x) + x\phi'(x)[/imath]
[imath]u''(x) = \phi'(x) + \phi'(x) + x\phi''(x) = 2\phi'(x) + x\phi''(x)[/imath]
Plugging in original equation.
[imath](1 + x^2)[2\phi'(x) + x\phi''(x)] + 2x[\phi(x) + x\phi'(x)] - 2x\phi(x) = 0[/imath]
[imath](1 + x^2)[2\phi'(x) + x\phi''(x)] + 2x^2\phi'(x) = 0[/imath]
[imath]2\phi'(x) + x\phi''(x) + 2x^2\phi'(x) + x^3\phi''(x) + 2x^2\phi'(x) = 0[/imath]
[imath](x^3 + x)\phi''(x) + (4x^2 + 2)\phi'(x) = 0[/imath]
I will add this extra magic to be able to use an integrating factor. Let [imath]m(x) = \phi'(x)[/imath]
[imath](x^3 + x)m'(x) + (4x^2 + 2)m(x) = 0[/imath]
[imath]\displaystyle m'(x) + \frac{(4x^2 + 2)}{(x^3 + x)}m(x) = 0[/imath]
[imath]\displaystyle m(x) = \frac{C}{e^{\int \frac{(4x^2 + 2)}{(x^3 + x)} dx}} = \frac{C}{x^4+x^2}[/imath]
We know that.
[imath]\displaystyle \frac{d\phi(x)}{dx} = m(x)[/imath], then
[imath]d\phi(x) = m(x) \ dx[/imath]
[imath]\displaystyle \int d\phi(x) = \int m(x) \ dx = \int \frac{C}{x^4+x^2} \ dx [/imath]
[imath]\displaystyle \phi(x) = -C\tan^{-1}x - \frac{C}{x} + D [/imath]
[imath]u(x) = \displaystyle x\phi(x) = -Cx\tan^{-1}x - \frac{Cx}{x} + Dx [/imath]
[imath]u(x) = -Cx\tan^{-1}x - C + Dx [/imath]
Combining the first solution [imath]u(x) = x[/imath] with this second solution, we get.
[imath]u(x) = -Cx\tan^{-1}x - C + Dx + x[/imath]
[imath]u(x) = -Cx\tan^{-1}x - C + Ex[/imath]
What's wrong with the negative sign??
Why do you think that there is something wrong with "negative" sign?
Hint: Look at post #1
If "2" is a constant - so is "-2".
I will not pretend to be an expert like Bob, but I don't agree with you. If the original solution has positive coefficients and your attempted solution has negative coefficients, isn't this an indication that something went wrong??
These are constants of "integration" - to be solved (determined) "by initial and or boundary conditions". C1 & C2 are ARBITRARY constants.
Calculate the expressions for u'(x) and u"(x) and use those in your original DE. If your DE is satisfied (and you did not make any algebraic/arithmetic mistake) - your solution is correct.
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Dr.Peterson
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The coefficients are neither positive nor negative. When you write -c, that is negative if c is positive, but positive if c is negative.
[imath]\displaystyle u(x) = -Cx\tan^{-1}x - C + Ex[/imath]
[imath]\displaystyle u'(x) = -C\tan^{-1}x - \frac{Cx}{x^2 + 1} + E[/imath]
[imath]\displaystyle u''(x) = - \frac{2C}{(x^2 + 1)^2}[/imath]
Original differential equation.
[imath]\displaystyle (1 + x^2)u''(x) + 2xu'(x) - 2u(x) = 0[/imath]
[imath]\displaystyle (1 + x^2)\left[- \frac{2C}{(x^2 + 1)^2}\right] + 2x\left[-C\tan^{-1}x - \frac{Cx}{x^2 + 1} + E\right] - 2[-Cx\tan^{-1}x - C + Ex] = 0[/imath]
[imath]\displaystyle - \frac{2C}{(x^2 + 1)^2} - \frac{2Cx^2}{(x^2 + 1)^2} - 2Cx\tan^{-1}x - \frac{2Cx^2}{x^2 + 1} + 2Ex + 2Cx\tan^{-1}x + 2C - 2Ex = 0[/imath]
[imath]\displaystyle - \frac{2C}{(x^2 + 1)^2} - \frac{2Cx^2}{(x^2 + 1)^2} - \frac{2Cx^2}{x^2 + 1} + 2C = 0[/imath]
[imath]\displaystyle - \frac{2C}{(x^2 + 1)^2} - \frac{2Cx^2}{(x^2 + 1)^2} + \frac{2C}{x^2 + 1} = 0[/imath]
[imath]\displaystyle - \frac{2C}{(x^2 + 1)^2} - \frac{2Cx^2}{(x^2 + 1)^2} + \frac{2Cx^2}{(x^2 + 1)^2} + \frac{2C}{(x^2 + 1)^2} = 0[/imath]
Confirmed.
What you were saying was true. The negative signs did not change the validity of the solution. Thank you professor Khan.
Thank you for passing by Dr.Peterson.
Let [imath]\displaystyle j = -C[/imath]
[imath]\displaystyle u(x) = jx\tan^{-1}x + j + Ex[/imath]
Would you consider the coefficient [imath]\displaystyle j[/imath] positive, negative, or neither?
Dr.Peterson
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Clearly, it's positive if C is negative, and negative if C is positive. So it's unknown!
When you obtain a solution involving an arbitrary constant C, it is still valid if you replace C with -C, or with 2C, or whatever; they are all equivalently arbitrary. (You should recall that in solving differential equations, we routinely "change the name" of a constant to make the answer simpler.)
I have never encountered a negative sign in the solution while solving differential equations. Therefore, I was tricked that I have done something wrong. (No one told me before.)
If I were un professeur et my students did a trivial mistake because I did not give them un exemple to cover that, I would never blame them.
Thanks Dr.Peterson for the clarification. This lesson deserves a noble prize! Or may be not.
Dr.Peterson
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So blame whoever taught you! (I didn't blame you, did I? But I was surprised, for good reason, that you would not be familiar with this.)
Here is one example of what I referred to, so you can see at least that it is taught somewhere:
8.3: Separable Differential Equations
We now examine a solution technique for finding exact solutions to a class of differential equations known as separable differential equations. These equations are common in a wide variety of …
math.libretexts.org
Now we use some logic in dealing with the constant [imath]C[/imath]. Since [imath]C[/imath] represents an arbitrary constant, [imath]3C[/imath] also represents an arbitrary constant. If we call the second arbitrary constant [imath]C_1[/imath], where [imath]C_1=3C[/imath],
the equation becomes
[math]\ln|3y+2|=x^3−12x+C_1.[/math]
Now exponentiate both sides of the equation (i.e., make each side of the equation the exponent for the base e).
[math]e^{\ln|3y+2|}=e^{x^3−12x+C_1}\\|3y+2|=e^{C_1}e^{x3−12x}[/math]
Again define a new constant [imath]C_2=e^{C_1}[/imath] (note that \(C_2>0):
[math]|3y+2|=C_2e^{x3−12x}.[/math]
And so on. We constantly rename constants.Here's another source that says more, first about negatives in particular:
Before equation (7), he says:
Now, from a notational standpoint we know that the constant of integration, c, is an unknown constant and so to make our life easier we will absorb the minus sign in front of it into the constant and use a plus instead. This will NOT affect the final answer for the solution.
Again, after equation (7), it says,
There is a lot of playing fast and loose with constants of integration in this section, so you will need to get used to it. When we do this we will always to try to make it very clear what is going on and try to justify why we did what we did.
In another place, he talks about similar manipulation, in the context of integrals:
In the first example, he says,
Now, both c and k are unknown constants and so the sum of two unknown constants is just an unknown constant and we acknowledge that by simply writing the sum as a c.
In the second example:
However, since the constant of integration is an unknown constant dividing it by 2 isn’t going to change that fact so we tend to just write the fraction as a c.
You are very resourceful Dr.Peterson. Although, I hate reading, I have read all the links that you have given me. One month ago, I thought I mastered differential equations. After this problem, I discovered that differential equations are an evil. Everyday, I learn something new. Now I have the full picture of the arbitrary constants. Thanks a lot professeur Peterson.