Initial Value Problem (with y(0) ?)

[kape]

Initial Value Problem

I (kind of) understand how to do initial value problems. I know that if the problem is $$y' = x, y(4) = 3$$ then you just differentiate it, solve for y(4) and then replace the constant C (in the equation I differentiated) with this answer.

But what if the initial value is y(0) = 6??

The problem I am trying to do is:

$$y' = 36 + y^2$$

$$y(0) = 6$$

$$0 < c < \frac{\pi}{2}$$

Differentiating, I get $$y = 36y + \frac{1}{3}y^3 + C$$.

But if I solve for y(0) = 6, then I get $$y = 36(0) + \frac{1}{3}(0)^3 + C$$.

But the answer can't be $$y = 36y + \frac{1}{3}y^3 + 6$$ - can it?!

And what is the significance of $$0 < c < \frac{\pi}{2}$$ - why is this needed?

I've been looking all over my calculus book, but I just don't understand how to do this. I am taking Engineering Mathematics (Kreyzig Book) and I can't help but feel that there is this large gap between this book and my calculus skills.. everything just looks so different.. *sigh*

Can anyone help me out?

 

[arildno]

Do you know what an independent variable is, and what it means to integrate with respect to that?

 

[kape]

No, I'm afraid not. Does it have anything to do with implicit differentiation?

 

[arildno]

x is your independent variable, y is your dependent.

Go back to your fallacious "integration" of the diff. eq.

On your left hand side, you have found the anti-derivative of y'(x) correctly, that is simply y(x).

But which variable have you integrated with respect to on your right hand side?
The dependent or the independent variable?

 

[kape]

Thank you for telling me about the independent and dependent variable. It now rings a bell.

I just realized that I said differentiation instead of integration.

As you can see, I am hopelessly confused..

I have only encountered equations where (for dy/dx) there are x's on the right hand side. I have never encountered equations with y's and x's and, as in this case, only y's.

I.. don't really know where to start. What should I be thinking of doing?

 

[HallsofIvy]

You are right about the discrepancy between what you are trying to do and your calculus skills! Plan to spend a lot of time reviewing calculus.

In the first place, you aren't "differentiating", you are "integrating". And the integral of y with respect to x is not (1/2)y²: y is some function of x and without knowing what function you don't know how to integrate it- that's the whole point of differential equations!

Actually, the part you say you are stuck on- that you are given y(0) rather than y(4) is trivial- just use x= 0 instead of x= 4! However, the part that you think you can do is completely wrong.

Your equation is y'= 36+ y². You might see it better if you wrote that as $$\frac{dy}{dx}= 36+ y^2$$. Separating into "differentials" $$dy= (36+ y^2)dx$$. Now can you see that the right hand side has "dx" rather than "dy"? You can't just treat that as an integral. In this simple example, a "separable" equation in which x and y can be completely separated, however, you can write it as
$$\frac{dy}{y^2+ 36}= dx$$.
Can you integrate that (think "arctan")?

Once you've done that integration and have y as a function of x, with the constant of integration, set x= 0 and y= 6 to determine the constant.

 

[kape]

My god. You are a GOD. Thank you SO much. I feel as a mortal would feel when a god demonstrates his powers by parting the sea or something. I am finally liberated from hours of trying and failing to understand my textbook.

I thank you. From the bottom of my heart. You don't know how happy I feel right now!

:D


So to continue, it becomes

$$\int \frac{dy}{y^2+6^2} = \int dx$$

$$\frac{1}{6}arctan\frac{y}{6} + C = x$$

Then

$$C = x - \frac{1}{6}arctan\frac{6}{6} = 0 - \frac{1}{6} . \frac{\pi}{4} = - \frac{\pi}{24}$$




But then.. if I do this, I come up with a different answer:

$$\frac{1}{6}arctan1 = x - C$$

$$arctan1 = 6x - 6C$$

$$arctan1 = 6(0) - C$$

$$C = -arctan1 = - \frac{\pi}{4}$$


What am I doing wrong?!

Also, if you cannot separate the variables, what do you do? Do equataions that are not separable have a separate name?

 

[arildno]

But then.. if I do this, I come up with a different answer:

$$\frac{1}{6}arctan1 = x - C$$

$$arctan1 = 6x - 6C$$

$$arctan1 = 6(0) - C$$ WHERE DID YOUR 6C TERM GO TO??

$$C = -arctan1 = - \frac{\pi}{4}$$


What am I doing wrong?!

Separable differential equations constitute a subset of differential equations.
Differential equations that are not separable are called..non-separable.

 

[HallsofIvy]

My god. You are a GOD. Thank you SO much. I feel as a mortal would feel when a god demonstrates his powers by parting the sea or something. I am finally liberated from hours of trying and failing to understand my textbook.

I thank you. From the bottom of my heart. You don't know how happy I feel right now!

:D

Woah! My sarcasm detector is going off big time!

 

[kape]

No, no, no! Though I am fairly sarcastic at times, I assure you I was NOT sarcastic this time! I realize that I did overdo it a little bit, but I really was overwhelmed with joy! You don't know how happy I was to be liberated from hours of frustration... and to be able to solve a question in this awful book for once! I also recognise you from helping me out a few days ago.. And most probably, I will need your help again in the future! (I have a very long and hard term ahead of me..) I don't think I'd be sarcastic, even if i wanted to - and this book has a way of humbling people like me. But seriously, I was, and still am very grateful! :)

As for the 6C.. I thought that anything multiplied with a constant, remains a constant. I've seen this in my calculus book.. Does it not apply in this situation? Though, yes, I now understand that it is that 6C that is the difference between the two versions..

 

[kape]

I also encountered something new and I'm not sure what to do:

The problem is:

$$y' = \frac{1-7y-49x}{1+y+7x}$$

with $$\inline y^2(1)=7$$ using $$\inline (y + 7x = v)$$

I've managed this so far:

$$v'-7 = \frac{1-7v}{1+v}$$

$$v' = \frac{-7(1+v)}{1+v} + \frac{8}{1+v} + 7$$

$$\frac{dv}{dx} = \frac{8}{1+v}$$

$$\int (1 + v) dv = \int 8 dx$$

$$v + \frac{1}{2}v^2 = 8x + C$$

But how do I plug that into $$(y = v - 7x)$$ to get the constant?!

How do I reduce $$v + \frac{1}{2}v^2 = 8x + C$$ for a single v?

 

[arildno]

Assuming you have calculated correctly,
solve for v(x) using the quadratic formula.

 

[kape]

Oh! But... doesn't a quadratic equation have to be in the form ax^2 + bx + c = 0; doesn't it have to equal zero?

How do I do it if it equals 8x?

Could you tell me about the 6C = C bit from the previous equation? I'm running into a lot of equations where I have to multiply a constant, and it is very important for me to know whether 6C = C or 6C = 6C (so that if it is 6C = 2, then C = 2/6)..

 

[arildno]

As for the "C"-thing:
Different C's are for y-functions with different initial conditions than yours, but fulfilling the same differential equation.

C is fixed for a particular solution of the diff. eq.

As for the quadratic formula:
Write this as:

$$\frac{1}{2}v^{2}+v+(B-8x)=0 (B=-C)$$

That is:
$$v=-1\pm\sqrt{1-2(B-8x)}$$
The constant B is determined from initial conditions.

 

[arildno]

I'm sorry, I've suddenly seen that you've made a total mess of evaluating your initial condition earlier. I'll post a reply later.

 

[arildno]

But then.. if I do this, I come up with a different answer:

$$\frac{1}{6}arctan1 = x - C$$

What ARE you doing here? x equals zero here!
So, you've got:
$$\frac{1}{6}arctan(1)=-C\to{C}=-\frac{\pi}{24}$$