Consider the differential equation dy/dx = 2x – y. (a) The slope field for the given differential equation is provided. Sketch the solution curve that passes through the point (l, 0) and sketch the solution curve that through the point (0, 1). (b) Find the value of b for which y = 2x + b is a solution to the differential equation. Justify your answer. (c) Let g be the function that satisfies the given differential equation with the initial condition g(0) = 0. Does the graph of g have a local extremum at the point (0, 0)? If so, is the point a local maximum or a local minimum? Justify your answer. (d) Show that if C is a constant, then y = Ce^-x + 2x – 2 is a solution of the differential equation
The correct answer and explanation is:
Here’s a detailed explanation and solution for each part of the problem:
(a) Sketching the solution curves
- Slope field: A slope field is a graphical representation of the slopes given by the differential equation dydx=2x−y\frac{dy}{dx} = 2x – y at various points in the xyxy-plane.
- Through (1,0)(1, 0):
- The slope at (1,0)(1, 0) is dydx=2(1)−0=2 \frac{dy}{dx} = 2(1) – 0 = 2.
- Sketch the curve starting at (1,0)(1, 0), following the slopes of the field. The curve should increase since the slope is initially positive.
- Through (0,1)(0, 1):
- The slope at (0,1)(0, 1) is dydx=2(0)−1=−1 \frac{dy}{dx} = 2(0) – 1 = -1.
- Sketch the curve starting at (0,1)(0, 1), decreasing initially because the slope is negative.
(b) Finding bb for y=2x+by = 2x + b to be a solution
Substitute y=2x+by = 2x + b into the differential equation dydx=2x−y\frac{dy}{dx} = 2x – y:
- Derivative: dydx=2\frac{dy}{dx} = 2.
- Substitution: 2=2x−(2x+b)2 = 2x – (2x + b).
- Simplify: 2=−b2 = -b, so b=−2b = -2.
Thus, y=2x−2y = 2x – 2 is a solution.
(c) Does gg have a local extremum at (0,0)(0, 0)?
The derivative is dydx=2x−y\frac{dy}{dx} = 2x – y. At (0,0)(0, 0), dydx=2(0)−0=0\frac{dy}{dx} = 2(0) – 0 = 0, so the slope is zero.
To determine if (0,0)(0, 0) is a local extremum:
- Find d2ydx2\frac{d^2y}{dx^2} by differentiating dydx=2x−y\frac{dy}{dx} = 2x – y: d2ydx2=2−dydx.\frac{d^2y}{dx^2} = 2 – \frac{dy}{dx}.
- At (0,0)(0, 0), dydx=0\frac{dy}{dx} = 0, so d2ydx2=2\frac{d^2y}{dx^2} = 2, which is positive.
Since d2ydx2>0\frac{d^2y}{dx^2} > 0, (0,0)(0, 0) is a local minimum.
(d) Verifying y=Ce−x+2x−2y = Ce^{-x} + 2x – 2 is a solution
Substitute y=Ce−x+2x−2y = Ce^{-x} + 2x – 2 into dydx=2x−y\frac{dy}{dx} = 2x – y:
- Compute dydx\frac{dy}{dx}: dydx=−Ce−x+2.\frac{dy}{dx} = -Ce^{-x} + 2.
- Substitution into dydx=2x−y\frac{dy}{dx} = 2x – y: −Ce−x+2=2x−(Ce−x+2x−2).-Ce^{-x} + 2 = 2x – (Ce^{-x} + 2x – 2).
- Simplify: −Ce−x+2=−Ce−x+2.-Ce^{-x} + 2 = -Ce^{-x} + 2.
Both sides are equal, confirming that y=Ce−x+2x−2y = Ce^{-x} + 2x – 2 satisfies the differential equation.
Explanation
This problem involves solving, verifying, and interpreting a first-order differential equation dydx=2x−y\frac{dy}{dx} = 2x – y. For part (a), the solution curves are sketched by analyzing the slope field provided, showing how the curve behaves starting at specific initial points.
In part (b), substituting y=2x+by = 2x + b reveals b=−2b = -2, indicating that the equation y=2x−2y = 2x – 2 satisfies the differential equation. For part (c), the analysis of the function gg at (0,0)(0, 0) involves determining if the point is a local extremum. By calculating d2ydx2\frac{d^2y}{dx^2} and finding it positive at (0,0)(0, 0), it is confirmed as a local minimum.
Finally, part (d) verifies the general solution y=Ce−x+2x−2y = Ce^{-x} + 2x – 2 by substitution. The calculation shows consistency between the differential equation and the given function, demonstrating it satisfies the equation for any constant CC. This showcases the methodical process of solving and analyzing differential equations, emphasizing the interplay between algebraic manipulation and calculus concepts like slope, curvature, and critical points.