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We have to find the extremal functions $y(x)$ of the integral $\int_0^1(y^2-(y')^2)dx$.

MY TRY:I used the formula $\frac{\partial F}{\partial y} - \frac{d}{dx}(\frac{\partial F}{\partial y'}) = 0$ and finally got $(y')^2+y^2=c$ but i could not get the ans as $y=A\cos x+B\sin x$.Thank you.

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    Actually, $y=\sqrt{c} \sin (x+\phi) \implies y'=\sqrt{c} \cos (x+\phi) \implies y^2+y'^2=c$ – 2017-01-19

2 Answers 2

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The differential equation you get is correct. Let's define $k^2:=c$, then we have

$$(y')^2+y^2=k^2\implies y'=\pm\sqrt{k^2-y^2}$$

This is a seperable equation, thus we have

$$\pm\int\frac{dy}{\sqrt{k^2-y^2}}=x+A$$

The integral on the left gives you an inverse sine function for the positive sign and an inverse cosine function for the negative sign. I'm going to go with the positive sign here, the negative is completely analogous.

\begin{align} \int\frac{dy}{\sqrt{k^2-y^2}}&=\frac{1}{k}\int\frac{dy}{\sqrt{1-\frac{y^2} {k^2}}}\\ &=\int\frac{du}{\sqrt{1-u^2}}\\ &=\arcsin(u)\\ &=\arcsin\left(\frac{y}{k}\right) \end{align}

Thus, we have

$$y(x) = k\sin(x+A)$$

This can be rewritten in the form

$$y(x) = C\sin(x)+D\cos(x)$$

using trigonometric identities.

When going with the negative sign, you get an inverse cosine function. You can then rewrite it in the same way using trig identities to arrive the general form that you seek.

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    Extremal function of an integral and stationary function of an integral are same? – 2017-01-19
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after the euler equation you should get $ y''+y=0$