You can identify points on a graph by looking at them. The x coordinate is the horizontal line. and the vertical line is the y axis
I ussually think of them as arrows telling me where to go.
if it is > you go ----->
if it is < you go <------
Answer:
it's B
Step-by-step explanation:
Trust me
(a) It looks like the ODE is
<em>y'</em> = 4<em>x</em> √(1 - <em>y </em>^2)
which is separable:
d<em>y</em>/d<em>x</em> = 4<em>x</em> √(1 - <em>y</em> ^2) => d<em>y</em>/√(1 - <em>y</em> ^2) = 4<em>x</em> d<em>x</em>
Integrate both sides. On the left, substitute <em>y</em> = sin(<em>t </em>) and d<em>y</em> = cos(<em>t</em> ) d<em>t</em> :
∫ d<em>y</em>/√(1 - <em>y</em> ^2) = ∫ 4<em>x</em> d<em>x</em>
∫ cos(<em>t</em> ) / √(1 - sin^2(<em>t</em> )) d<em>t</em> = ∫ 4<em>x</em> d<em>x</em>
∫ cos(<em>t</em> ) / √(cos^2(<em>t</em> )) d<em>t</em> = ∫ 4<em>x</em> d<em>x</em>
∫ cos(<em>t</em> ) / |cos(<em>t</em> )| d<em>t</em> = ∫ 4<em>x</em> d<em>x</em>
Since we want the substitutiong to be reversible, we implicitly assume that -<em>π</em>/2 ≤ <em>t</em> ≤ <em>π</em>/2, for which cos(<em>t</em> ) > 0, and in turn |cos(<em>t</em> )| = cos(<em>t</em> ). So the left side reduces completely and we get
∫ d<em>t</em> = ∫ 4<em>x</em> d<em>x</em>
<em>t</em> = 2<em>x</em> ^2 + <em>C</em>
arcsin(<em>y</em>) = 2<em>x</em> ^2 + <em>C</em>
<em>y</em> = sin(2<em>x</em> ^2 + <em>C </em>)
(b) There is no solution for the initial value <em>y</em> (0) = 4 because sin is bounded between -1 and 1.
