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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.

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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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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