Answer
A) As $x→\infty, y'→\infty$ when $x $ is $ \infty$, the graph tends to be vertical
B) $y' = xy^{3}$
C) As $x→\infty, y'→\infty$ and $y' →0 $ as $x→0$.
The graph tends to be horizontal when $x→0$ and vertical when $x→\infty$
D) $y = (\frac{1}{4}- x^{2})^{-1/2}$
Work Step by Step
A) When x is close to 0, $y' = xy^{3}$ xy' is also close to 0
So $ y'→0$ as $x→0$
This shows that the slope of the graph tends to be horizontal as x tends to 0
As $x→\infty, y'→\infty$ when $x $ is $ \infty$, the graph tends to be vertical
B) $y = (c - x^{2})^{-1/2}$
$→ y' = -\frac{1}{2}(c -x^{2})^{-3/2}(-2x)$
$→ y' = x (c-x^{2})^{-3/2}$
$→y' = x [(c-x^{2})^{-1/2}]^{3} = xy^{3}$
Hence, $y = (c - x^{2})^{-1/2}$ is a solution of the equation $y' = xy^{3}$
D) Since $y'= xy'$ and $y(0)=2$
The solution of this equation is $y=(c-x^{2})^{-1/2}$
Substitute x = 0 and y=2
$(c-0)^{-1/2} = 2$
$\frac{1}{\sqrt c} = 2$
$→\sqrt c = \frac{1}{2}$
$→ c = \frac{1}{4}$
Hence, $y = (\frac{1}{4}- x^{2})^{-1/2}$
