Question:
Consider the differential equation \(\frac{dy}{dx}=2x – y.\)
(a) On the axes provided, sketch a slope field for the given differential equation at the six points indicated.
(b) Find \(\frac{d^{2}y}{dx^{2}}\) in terms of x and y. Determine the concavity of all solution curves for the given differential equation in Quadrant II. Give a reason for your answer.
(c) Let y = f(x) be the particular solution to the differential equation with the initial condition f (2) = 3.
Does f have a relative minimum, a relative maximum, or neither at x = 2 ? Justify your answer.
(d) Find the values of the constants m and b for which y = mx + b is a solution to the differential equation.
Answer/Explanation
Ans:
(b)
\(\frac{dy}{dx}= 2x – y\)
\(\frac{d^{2}y}{dx^{2}}= 2-\frac{dy}{dx}=2-(2x-y)\)
\(\frac{d^{2}y}{dx^{2}}= 2-2x+y\)
In Quadrant II, x < 0 and y > 0, So \(\frac{d^{2}y}{dx^{2}}= 2-2x+y>0,\)
Thus all solution curves in Quadrant II are concave up.
(c)
\(\frac{dy}{dx}= 2x-y=2.2-3 =1\)
Neither, as \(\frac{dy}{dx}\) ≠ 0 at x = 2.
(d)
\(\frac{dy}{dx}= 2x-y\), y = mx + b
\(\frac{dy}{dx}= m = 2x – y\)
m = 2x – (mx + b)
m = (2-m)x – b, equate coefficients
2 – m = 0
m = 2
-b = m
b = -m = -2.
m = 2, b = -2