# Homework Help: Initial Value ODE

1. Dec 15, 2013

### freezer

1. The problem statement, all variables and given/known data

Solve the initial value problem:

$$x^{2}{y}'' + x{y}' + y = 0, x>0, y(1)=1, {y}'=2$$

2. Relevant equations

y=x^m

3. The attempt at a solution

$$x^{2}(m(m-1)x^{m-2})+xmx^{m-1} + x^{m}$$

$$x^{2}(m(m-1)x^{m-2})+xmx^{m-1} + x^{m}$$

$$x^{m}(m(m-1) + m + 1)$$

$$m = \pm i$$

This is the way i was doing it:
$$C_1 e^{it} + C_2 e^{-it}$$

$$C_1(cos(t) + i sin(t)) + C_2(cos(t) - i sin(t))$$

The solution shows:
$$C_1 x^{i} + C_2 x^{-i}$$
$$C_1(cos(ln(x)) + i sin(ln(x))) + C_2(cos(ln(x)) - i sin(ln(x)))$$

With the initial conditions indicate that the solution is correct. Yet the text book shows the form to be:

$$Y(x) = C_1 e^{r_1 x} + C_2 e^{r_2 x}$$

not

$$Y(x) = C_1 x^{r_1} + C_2 x^{r_2}$$

And if the second form is correct, I have not found an Euler identity that supports the step.

2. Dec 15, 2013

### Staff: Mentor

Your assumption was that y = xm was a solution. Using that assumption, you solved for m and got ±i.
I'm confused. You say that
and then you say
Which of these does your book show as the solution?

If the solution you found (i.e., y = c1xi + c2x-i) satisfies the differential equation, then it is the general solution. From the initial conditions you can find the constants c1 and c2.

3. Dec 15, 2013

### vanhees71

You have the correct solution. From your findings of two linearly independent solutions
$$y_1(x)=\exp(\mathrm{i} \ln x), \quad y_2(x)=\exp(-\mathrm{i} \ln x).$$
The general solution is given by
$$y(x)=C_1 y_1(x) + C_2 y_2(x).$$
You can also choose any other two linearly independent solutions, e.g.,
$$\tilde{y}_1(x)=\frac{1}{2} [y_1(x)+y_2(x)]=\cos(\ln x), \quad \tilde{y}_1(x) = \frac{1}{2 \mathrm{i}} [y_1(x)+y_2(x)]=\sin(\ln x).$$
The general solution is again given by all linear combinations of these two functions.

4. Dec 15, 2013

### freezer

Okay, where does the ln x come from?

5. Dec 15, 2013

### freezer

The textbook shows in the examples the e^{rt} while the solution to this problem shows x^r and there is nothing in the text of the chapter that shows this process. I was able to take cos(ln(x) + i sin(ln(x) back to x^i in wolfram but not sure what i am missing where the ln(x) is coming from and the identity for the x^i.

6. Dec 15, 2013

### Staff: Mentor

xm = (eln(x))m = em*ln(x). This holds as long as x > 0.

7. Dec 15, 2013

### freezer

Sure e^{ln(x)} = x, but where did the ln(x) come from. I have went through the chapters in the book and I have not found a single theorem or example that shows this.

The examples from the chapter show that when you get your zeros, you use e^{rt). In this case in place of t they used ln(x). I just don't see where the ln(x) came from or what basis this was done with this problem.

Even over at Pauls notes does not show this:
http://tutorial.math.lamar.edu/Classes/DE/ComplexRoots.aspx

Last edited: Dec 15, 2013
8. Dec 15, 2013

### LCKurtz

Perhaps your text or teacher has talked about how the equation $ax^2y''+bxy' + cy = 0$ can be converted to a similar constant coefficent equation $a\ddot y + b\dot y + cy = 0$ by the change of variable $t=\ln x$. Then you solve the second equation for y as a function of $t$ with your usual $e^{rt}$ method. Once you have that solution you put $t=\ln x$ in and presto, you have the solution to the original problem. The method you describe sounds like that is what they are doing.

9. Dec 16, 2013

### ehild

You did not specify what t is. y is function of x. You assumed the solutions in form $$y=x^m$$ and you got

$$m = \pm i$$

The solutions are y1=xi and y2=x-i, the general solution is y=C1xi+C2x-i.

$$x=e^{\ln(x)}$$.

With that, you can write the solution also in the form
$$y=C_1 e^{i\ln(x)}+ C_2 e^{-i\ln(x)}$$

ehild

Last edited: Dec 16, 2013
10. Dec 16, 2013

### LCKurtz

I noticed a typo after it was too late to edit. The equation $a\ddot y + b\dot y + cy = 0$ should have been $A\ddot y + B\dot y + Cy = 0$. The transformed equation has different constants than the original.