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I am self studying ode from boyce diprima book and while doing exercises on chapter 2.6.

I couldnt undestand how question 17 is derived I checked the solution manual and I found the same answer which I have already found by myself but what the the question asks is different.

The integral on solution manual is not definite and the question asks me to show how a definite integral is derived and I am not sure exactly how that definite integral is derived

M(s,y0) : this expression on question makes me suspicious

I uploaded the question to scribd because it is too long to write together with the theorem they refer to.

http://www.scribd.com/fullscreen/77494675?access_key=key-1fqwhkmcex23ufgyz0sh

can anyone show me how the definite integral, question asks, can be derived ? I also also put the solution manuals answer to pdf.

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1 Answer 1

I'm pretty shaky with DEs myself, and the notation always confuses the heck out of me, but nobody else has answered this yet, so I thought I'd take a stab at it. Hopefully my answer will at least give you something to work with (or encourage better answers):

The $\int_{x_0}^{x} M(s, y_0) \, ds$ is probably just changing the name of the first variable inside $M$'s parenthesis. It is used to get get the result in the correct letter, $x$, after integrating.

Note also that the book probably uses the notation that $x$ is a variable, and $x_0$ is a fixed unknown (a.k.a. constant).

Now let me change notations here for a minute. Have you seen where sometimes they represent a function with a lowercase letter and the function's integral with an uppercase letter (e.g., $M'(x, y) = m(x, y)$)? I'm going to use that notation and try to find $\int_{x_0}^{x} m(s, y_0) \, ds$.

If the integral of $m(s, y_0)$ is $M(s, y_0)$ (that is, $\int m(s, y_0) \, ds = M(s, y_0)$), then

$$\int_{x_0}^{x} m(s, y_0) \, ds = \left. M(s, y_0) \right|_{x_0}^{x} = M(x, y_0) - M(x_0, y_0)$$

Since the last term, $M(x_0, y_0)$, is a number (aka constant), taking the derivative of this with respect to (wrt) $x$ yields $m(x, y_0)$.

Next, in the textbook's answer they used the definite integral of $N$ to get the value for $\psi$. That is if the partial derivative of $\psi$ wrt $y$ equals $N$ (e.g. $\frac{\partial \psi}{\partial y} = N$), then

$$\psi(x, y) = \int_{y_0}^{y} \frac{\partial \psi}{\partial y} \, dy = \int_{y_0}^{y} N(x, t) \, dt = \int N(x, y) \, dy - \int N(x, y_0) \, dy$$

The first term on the right hand side (rhs), $\int N(x, y) \, dy$, is written that way because we don't know the integral of $N$ (it's the capital of capital $n$, or a really big $N$ :/ ). It's not an indefinite integral in the usual sense, it's just a notation to indicate "the function that you'd get when you integrate $N(x, t)$ with respect to $t$ and then evaluate that function at $t = y$".

The second term, $\int N(x, y_0) \, dy$, is obtained the same way and it means "the function you get when you integrate $N(x, t)$ wrt $t$ and then evaluate the function at $t = y_0$". It is a function of $x$, and I think they just renamed it $h$, so that $h(x) = \int N(x, y_0)$.

Next, they took the partial derivative of $\psi$ wrt $x$ and set it equal to $M$ and solved for $h'(x)$. I don't understand why they reversed the integral and partial derivative signs, but I guess that's allowed, and I don't understand the rest of the book's answer from this point on. But I hope I at least clarified why the answer looked like it was using indefinite integrals. Maybe someone can now chip in with an explanation of the rest of the text's answer.

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Welcome to math.SE. You can typeset mathematics using MathJax by enclosing LaTeX code in $ or $$. –  Zhen Lin Jan 8 '12 at 3:06

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