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I am stuck with the following problem:

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I have to determine which of the aforementioned options is/are correct?

MY ATTEMPT: Using Lagrange's method I get , $$\frac {dx}{x}=\frac {dy}{y}=\frac {du}{2u}$$ so that $$\frac {dx}{x}=\frac {dy}{y} \implies x=ay\,\,$$ and $$\frac {dx}{x}=\frac {du}{2u} \implies x^2=bu\,\,\,$$,where $a,b$ are constants. Now,I am stuck .How should I use the initial conditions effectively . Can someone explain me with some details how to tackle this type of problems without going into geometrical interpretation of the problem? Thanks and regards to all.

learner
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2 Answers2

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You have: $$x^2 = bf(x),\ x = axg(x)$$ So for part 1): $$x^2 = 2bx,\ x = ax$$ There is obviously no constant $b$ such that $x^2=2bx$ fo all $x$, therefore there are no solutions.

Nathaniel Bubis
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  • ok. Got your point.Using the same logic for option 2,I get$,,a=1,b=1/2$. But how can I determine whether there is a unique solution or infinite no. of solutions. There may be more than one options true. So,how I check the other options keeping in mind whether there are unique /inf. no. of solns. can you clarify a bit more in this regard? – learner Jun 05 '13 at 17:02
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From your solutions wrote in a more convenient manner:

$$\frac{y}{x}=c_1$$

$$u=c_2x^2$$

But $c_2=h(c_1)$ With $h$ any single variable function. So

$$u(x,y)=x^2h(y/x)$$

Setting the initial conditions, $y/x=g(x)$ $u=f(x)$ we have:

$$f(x)=x^2h(g(x))$$

1.- $\mathbf{f(x)=2x, g(x)=1}$

$2x=x^2h(1)$; $2/x=h(1)$ for all $x$. So it's a intial condition impossible to satisfy. No solution.

2.- $\mathbf{f(x)=2x^2, g(x)=1}$

$2x^2=x^2h(1)$; $2=h(1)$. This initial condition is fulfilled by a whole family of functions, that with this constraint on one only value. Infinite number of solutions.

3.- $\mathbf{f(x)=x^3, g(x)=x}$

$x^3=x^2h(x)$; $h(x)=x$ Wich leads unambiguously to $u(x,y)=xy$. Unique solution.

4.- $\mathbf{f(x)=x^4, g(x)=x}$

$x^4=x^2h(x)$; $h(x)=x^2$ Leadig to $u(x,y)=y^2$. Unique solution.

Rafa Budría
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