Solve the Given Differential Equation by Separation of Variables

**Question 4** Consider the curve given by \(x^2 + 4y^2 = 7 + 3xy\). (a) Show that \(\frac{dy}{dx} = \frac{3y-2x}{8y-3x}\) (b) Show that there is a point P with y-coordinate 3 at which the line tangent to the curve at P is horizontal. Find the x-coordinate of P. (c) Find the value of \(\frac{d^2y}{dx^2}\) at the point P found in part (b). Does the curve have a local maximum, a local minimum, or neither at the point P? Justify your answer. **Solution:** (a) 2x + 8yy' = 3y + 3xy' (8y - 3x)y' = 3y - 2x y' = (3y - 2x) / (8y - 3x) (b) 3y - 2x = 0; 8y - 3x = 0 When x = 3, y = 2 3^2 + 4 * 2^2 = 25 and 7 + 3 * 3 * 2 = 25 Therefore, P = (3, 2) is on the curve and the slope is 0 at this point. (c) d²y/dx² = [(8y - 3x)(3y' - 2) - (3y - 2x)(8y' - 3)] / (8y - 3x)^2 At P = (3, 2), d²y/dx² = (16 - 9)(-2) / (16 - 9)^2 = -2/7 Since y' = 0 and d²y/dx² < 0 at (3, 2), the curve has a local maximum at P. **Question 5** Consider the differential equation: dy/dx = (3-x)/y (a) Let y = f(x) be the particular solution to the given differential equation for 1 < x < 5 such that the line y = -2 is tangent   to the graph of f. Find the x-coordinate of the point of tangency, and determine whether f has a local maximum, local minimum, or neither at this point.   Justify your answer. (b) Let y = g(x) be the particular solution to the given differential equation for   -2 < x < 8, with the initial condition g(6) = -4. Find y = g(x).   **Solution:** (a) dy/dx = 0 when x = 3 d^2y/dx^2 = -y - y'(3-x) / y^2 = 1/2 at (3, -2) So, f has a local minimum at x = 3. Because f is continuous for 1 < x < 5, there is an interval containing x = 3 on which y < 0. On this interval, dy/dx is negative to the left of x = 3 and positive to the right of x = 3. Therefore, f has a local minimum at x = 3. (b) y dy = (3 - x) dx (1/2)y^2 = 3x - (1/2)x^2 + C 8 = 18 - 18 + C; C = 8 y^2 = 6x - x^2 + 16 y = -sqrt(6x - x^2 + 16) AP Calculus BC-6 Consider the differential equation given by d/dx (y) = x * (y - 1) ^ y (a) On the axes provided, sketch a slope field for the given differential equation at the eleven points indicated. (b) Use the slope field for the given differential equation to explain why a solution could not have the graph shown below. (c) Find the particular solution y = f(x) to the given differential equation with the initial condition f(0) = - 1 (d) Find the range of the solution found in part (c). (a) + (b) The graph does not have slope 0 where y = 1 of The slope field shown suggests that solutions are asymptotic to y = 1 from below, but the graph does not exhibit this behavior. (e) 1/((y - 1) ^ 2) * dy = xdx - 1/(y - 1) = 1/2 * x ^ 2 + C 1/2 = 0 + C; C = 1/2 - 1/(y - 1) = 1/2 * (x ^ 2 + 1) y = 1 - 2/(z ^ 2 + 1) (d) range is - 1 <= y < 1 5 1: zero slope at 7 points with y = 1 and 0 21: negative slope at (-1,0) and (-1,-1) positive slope at (1.0) and (1,-1) steeper slope at y = - 1 than y = 0 1: reason 1: separates variables 1 antiderivativen 1: constant of integration 1: uses initial condition f(0) = - 1 1: solves for y 0/1 if y is linear Note: max 2/5 (1-1-0-0-0) if no constant of integration Note: 0/5 if no separation of variables 1: answer 0/1 if-1 not in range Question 5 Consider the differential equation 2-4 (a) The slope field for the given ditferential equation is provided. Sketch the solution curve that passes through the point (0, 1) and sketch the solution curve that passes through the point (0-1). (b) Lethe the function thai satisfies the given differential equation with the initial condition f(0)=1. Use Euler's method, starting at with a step size of 0.1, to approximate (0.2). Show the work that leads to your answer (c) Find the value of & for which y=2x+ is a solution to the given differential equation. Justify your answer. (d) Let be the function that satisfies tlin given differential equation with the initial condition 9(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. (a) (b) f(0.1)/(0)+(0)(0.1) 1+(2-0)(0.1) = 1.2 (0.2) (0.1)+(0.1) (0.1) 1.2+(2.4-0.4) (0.1) = 1.4 (c) Substitute y=2x+ in the DE: 2=2(2+6)-4 = 26, so b=1 OR Guess & 1, y=2x+1 (4x+2) Verify: 2-4( 0-4-2 (d) g has local maximum at (0,0). '(0)= dy = = 2(0)-4(0) dy 2 de 40-4 (0) 2 = (0)"و. -0, and 80 2 2 1: solution curve through (0,1) 1: solution curve through (0,-1) Curves must go through the indicated points, follow the given slope lines, and extend to the boundary of the slope field. 1: Euler's method equstions or equivalent table applied to (at least) two iterations 1: Euler approximation to f(0.2) (not eligible without first point) d 1: uses(2x+b)=2 in DE 11 2 1: g'(0)=0 31: shows g(0) = -4 1: conclusion