Question

Shown above is a slope field for which of the following differential equations? 

(A) \(\frac{d y}{d x}=\frac{x(4-y)}{4}\)

(B) \(\frac{d y}{d x}=\frac{y(4-y)}{4}\)

(C) \(\frac{d y}{d x}=\frac{x y(4-y)}{4}\)

(D) \(\frac{d y}{d x}=\frac{y^{2}(4-y)}{4}\)

Answer 1

Two things need to be noted instantly from the graph

The slope is parallel to \(x\) -axis at \(y=0 \& y=4\)

\(y=0 \Longrightarrow \frac{d y}{d x}=0\)

This rules out (A)

since

Then \(y=0 \Rightarrow \frac{d y}{d x}=\frac{4 x}{4}=x \neq 0 \quad \forall x\)

Next , we observe in the sketch that

\(\frac{d y}{d x} \neq 0 \quad x=0\)

This rules out (C) since then

\(x=0 \Longrightarrow \frac{d y}{d x}=\frac{0 y(y-4)}{4}=0 \quad \forall y\)

Hence \(\mathrm{B} \& \mathrm{D}\) are the only candidates.

Next the slope changes sign at \(y=4\). However, since \(y>0\) at \(y=4\). Both B\& D satisfy the condition. Thus this argument can't be used to leminate the other.

Now notice at \(y=0\) i.e. near \(x\) -axis the slope is same for \(y<0 \& y>0\).

Hence if we apply Rolles theorem for

\(g(x, y)=\frac{d y}{d x}\)

we find

\(g\left(x, 0_{-}\right)=g\left(x, 0_{+}\right)\)

Hence there exists a point near \(\mathrm{y}=0\) whereby

\(\frac{\partial g}{\partial y}=\frac{\partial}{\partial y}\left(\frac{d y}{d x}\right)=0 \quad y \sim 0\)

For option (B)

\(\frac{\partial g}{\partial y}=\frac{\partial}{\partial y}\left(\frac{d y}{d x}\right)=\frac{4-2 y}{4} \Longrightarrow \lim _{y \rightarrow 0} \frac{\partial g}{\partial y}=1 \neq 0\)

Where as for option (D)

\(\frac{\partial g}{\partial y}=\frac{\partial}{\partial y}\left(\frac{d y}{d x}\right)=\frac{8 y-3 y^{2}}{4} \Longrightarrow \lim _{y \rightarrow 0} \frac{\partial g}{\partial y}=0\)

Hence (D) is correct option