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While working through the exercises in a book on the Calculus of Variations, I've hit a roadblock in trying to solve this differential equation: $C_1=4y^2+(y')^2+8y$

Let me back-up a bit and fill-in the larger context. The book I'm using is "Calculus of Variations", by Lev D. Elsgolc (Dover, ISBN13: 978-0-486-45799-4), and the above equation results from an intermediate step in problem 6 of chapter 3.

The task set forth in the problem is to examine the extrema of the functional $v(y(x))=\int_0^{\pi/4}(4y^2-(y')^2+8y)dx$, where $y(0)=-1$ and $y(\pi/4)=0$.

Since the functional $F(y,y')=4y^2-(y')^2+8y$ is not dependent on $x$, rather than using the general form of the Euler equation, I can instead use the special case in which $C_1=F-y'F_{y'}$ (c.f., pp 36). This is how I obtained the differential equation at the beginning of my question. So far I've tried various methods to solve it, but without success.

For some of the more complicated examples in previous chapters, the author has used trig substitutions to express the equation in terms of a different parameter/variable, which is easier to solve. However, this approach hasn't worked here.

Another approach that has yielded little fruit is to substitute the general form solution $y=A e^{ax}+Be^{bx}$. But this just seems to yield too many unknowns in too few equations.

I've even tried the naive approach of looking through the Schaum's outline on differential equations to see if there are any exercises similar to my problem. It's not that I want a mindless recipe to apply blindly, I just want to find a clue as to what chapter in that book would cover solutions to an equation of this form.

It certainly looks like a first-order equation to me. But I've seen no examples covering the case in which the first derivative term has an exponent.

I'll admit that I'm very rusty at differential equations, and haven't really had much practice at them since engineering school (15-20 years ago).

Am I missing something simple or elementary? If someone would be able to point me in the right direction, that would be a great help. If the right answer is to go back to my differential equation textbook and start from the beginning, then that's what I'll do.

Incidentally, I've marked this with the "homework" tag, but this is entirely self-study, and not for any class that I'm enrolled in. But since this pertains to an end-chapter problem in a textbook, I played it safe and tagged it as homework anyway.

Update: I probably should have mentioned this originally in my problem description, but the book provides the solution to this problem: a strong maximum at $y=sin(2x)-1$ Not that this omission was an obstacle to those who did respond, but I figured I should clarify that I have the solution, but I was having trouble with the intermediate steps towards reaching it.

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Is $C_1$ a constant? –  user17762 Jan 14 '13 at 6:45
    
Yes, it is a constant. I should have specified that. Also, I should note that I'm just sticking with the author's convention of putting a subscript on the constant. –  PaulDeLong Jan 14 '13 at 6:50
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2 Answers

up vote 3 down vote accepted

We have $$(y')^2 = C_1 - 8y - 4y^2 = C_1+4 - 4(y+1)^2$$ Let us call $C_1 + 4$ as another constant say $c^2$. We then get that $$y' = \sqrt{c^2 - 4(y+1)^2} \implies dx = \dfrac{dy}{\sqrt{c^2 - 4(y+1)^2}}$$ Now setting $y+1 = \dfrac{c}2 \sin(t)$, we get that $dy = \dfrac{c}2 \cos(t) dt$ and $c^2 - 4(y+1)^2 = c^2 \cos^2(t)$. Hence, $$dx = \dfrac{\dfrac{c}2 \cos(t) dt}{c \cos(t)} = \dfrac{dt}2 \implies x = \dfrac{t}2 - k$$ Hence, we get that $$x+ k = \dfrac{\arcsin(2(y+1)/c)}2 \implies \dfrac{2(y+1)}c = \sin(2x+ 2k)$$ This gives us $$y = -1 + \dfrac{\sqrt{C_1+4}}2 \sin(2x+\phi)$$ where $\phi$ is a constant.

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Thanks. And once I use the boundary conditions, I'm able to arrive at the given solution, using $C_1=0$ and $\phi = n\pi$. I'm now working on finding-out why this particular function is a strong maximum, but I didn't specify that as part of the scope of my original question. So thanks again for your help. –  PaulDeLong Jan 17 '13 at 19:30
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If you differentiate your whole equation once you get:

$$0 = 8yy'+2y'y''+8y'= 2y'(y''+4y+4)$$

and you can find your solution from the linear equation

$$y''+4y+4 = 0$$

There must be a deep reason for this, but the "differentiate everything once more" works very often when dealing with calculus of variations.

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Oh cool, I actually tried that; but for reasons I can't adequately remember, I gave-up on it. I'll have to pursue that and see where it takes me. Thanks. –  PaulDeLong Jan 14 '13 at 7:23
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I did get around to working-out this approach to its conclusion. I had forgotten the basic technique for solving an equation of this form, so I had to go brush-up on it. And I did arrive at the right answer. I'm sure you don't need the play-by-play, but I'll post it here anyway for posterity. In broad strokes, using $y_h = C_1 \exp(+2ix) + C_2 \exp(-2ix)$ for the homogeneous solution, and $y_p = C_3$ as the particular solution, and using the initial conditions described earlier, I arrived at constants $C_1=-i/2, C_2=+i/2, C_3=-1$, and the full solution becomes $y=y_h+y_p=sin(2x)-1$ –  PaulDeLong Jan 17 '13 at 19:04
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