# A difficulty in understanding the solution of Exam GRE 0568 Q31.

The question and its answer is given in the following picture: The question was asking about "the graph of a solution to the differential equation", so why in the solution he is speaking about $dy/dx$ and not speaking about $y$? could anyone explain this for me please?

Also I do not understand why our selection is narrowed to only (A) and (B) and not to (A)(B)(D) and (E), why the author exclude (D) and (E)? could anyone explain this for me?

Finally, it is not clear for me why he exclude (B), could anyone explain this for me please?

See $\frac {dy}{dx}$ is the slope of the curve at that point. Now as $y\ to \pm \infty$ the slope goes to $\infty$ according to the equation. So the slope of graph at both the infinities should be $\infty$. This is true only for $A,B$. Now see graphs around $0$ we have slope as $1$ around $0$ according to the equation. But in $B$ the sope around $x=0$ is almost $0$ thus the correct graph is $A$ as its increasing around $0$.

The equation $y'=1+y^4$ tells you information about slopes of tangent lines to the graph of $y=y(x).$ Since this is a GRE exam problem, here is how I would do it.

Since $1+y^4\geq 1$ for all $y$, by just looking at the slopes, one can tell immediately that B, C, D, E are all wrong. Why? All the graphs in those choices have horizontal or almost horizontal tangent lines, which contradicts the assumption that $y'\geq 1$ for all $y$.

Choose A.

• So smart solution. – user426277 Jun 19 '17 at 20:31

First thing we notice is that the derivative must always be positive and is bounded by 1, as $1+y^4\geq 1$. Because of this, you can exclude case E, as the derivative is negative on the left part, and B, as the derivative on this graph has value 0 at 0 (flat), while our formula would give us a derivative at least 1. The cases C and D also clearly dont respect that the derivative must be at least 1 : for C the function flattens when going to plus or minus infinity, while D flattens when going to minus infinity.

I'd like to note that i assumed here that the graphs have representative axis, as my argument might not work if the axes had greatly different scales.

• "and B, as the derivative on this graph has value 0 at 0 (flat)" why did you say this, the graph is higher than 0 at 0.? I feel that you are using the idea that the derivative is the slope.... right? – user426277 Jun 19 '17 at 14:42
• Exactly. At $x=0$, the graph slope is flat, and thus the derivative is 0. – tjeremie Jun 19 '17 at 15:00

They speak about the graph of a solution, but it is then found easy to speak about a particular apsect of said graph, i.e. the slope of tangent lines to it.

From your ODE $$\frac{\mathrm{d}y}{\mathrm{d}x} = 1 + y^4$$ they conclude that $$\frac{\mathrm{d}y}{\mathrm{d}x} \to \infty$$ as the modulus of $y$ goes to infinity.

This means, in plain english, the slope goes to (positive) infinity as the value of the function goes to positive or negative infinity.

So let us consider each graph under question:

A) as $y$ grows in both sense, the slope goes to positive infinity: compliant with the condition previously derived

B) same as above, but the behaviour around $x=0$ is now of interest. The value $y(0)$ is different from zero and is positive: from the ODE the derivative cannot be zero (this would lead to the impossible equality $0$ = $1$ + positive value

D) for $y$ approaching zero, the slope has to approach 1, as one verifies from the ODE. Both $y$ and its derivative cannot tend to zero, according to the ODE (in the limit it would read $0 = 1 + 0$)

E) as $y$ grown for negative $x$, the slope is negative, contradditicting the requirement noted above.