Applied Differential Equations. Midterm Exam 1

Question 1. [18 points] Determine if the following statements are TRUE or FALSE. Justify each answer. (a)[3 points] The equation y 000 (1 + y) = 2t is a third order linear ODE. Solution: False. The equation is not linear. It cannot be written in the form a 3 (t )y 000 + a 2 (t )y 00 + a 1 (t )y 0 + a 0 (t )y = g (t ). (b)[3 points] The function y(x) = e −2x is a solution for x 2 y 00 − 2x y 0 + 5y = 0. Solution: False. y(x) = e −2x , y 0 (x) = −2e −2x , y 00 = −2(−2)e −2x = 4e −2x . x 2 y 00 − 2x y 0 + 5y = x 2 (4e −2x ) − 2x(−2e −2x ) + 5e −2x = 4x 2 e −2x + 4xe −2x + 5e −2x = (4x 2 + 4x + 5)e −2x which is always nonzero. (c)[3 points] The equation x y 0 = (x 2 + 3)(y 2 + 1) is a separable ODE. dy Solution: True. x y 0 = (x 2 + 3)(y 2 + 1) ⇒ x d x = (x 2 + 3)(y 2 + 1). We can separate variables to have dy y 2 +1 = x 2 +3 d x. x (d)[3 points] dy + p(x)y = q(x) cannot be solved if p(x) or q(x) are discontinuous. dx Solution: False. Since there exist many examples where solutions exist. (e)[3 points] The isoclines of an autonomous ODE, dy = f (y), are always horizontal. dx dy Solution: True. Since autonomous ODE d x = f (y). The isoclines of it: f (y) = C does not depend on x. The slope does not depend on x in an autonomous ODE. (f)[3 points] The linear differential equation, y 0 + k 1 y = k 2 , where k 1 and k 2 are nonzero constant, always possesses a constant solution. Solution: True. Rewrite the equation of y 0 + k 1 y = k 2 to y 0 = k 2 − k 1 y. By setting y 0 = 0, i.e., k 2 −k 1 y = 0 =⇒ y = kk21 since (k 1 & k 2 are nonzero constant). Then y = kk21 is the constant solution. 2 Question 2. [10 points] Consider the following differential equation dy p = xy dx determine a region of the x y-plane for which the given differential equation would have a unique solution whose graph passes through a point (x 0 , y 0 ) in the region. Solution: Let f (x, y) = • f (x, y) = • p p x y, then x y is continuous when x ≥ 0, y ≥ 0 or x ≤ 0, y ≤ 0; 1 1 ∂f 1 x = (x y) 2 −1 · x = p ∂y 2 2 xy Therefore, by the Fundamental Theorem of Existence & Uniqueness, in the region of R = {(x, y)|x > 0, y > 0, or x < 0, y < 0} would have a unique solution whose graph passes through a point (x 0 , y 0 ) in the region. 3 Question 3. [30 points] Consider the first order autonomous differential equation dy = y 2 (9 − y 2 ) dt Answer the following questions without solving the equation. (a) [6 points] Denote f (y) = y 2 (9 − y 2 ), sketch the graph of f (y) versus y; Solution: The graph of f (y) versus y is given as follow 50 40 30 20 f(y) 10 0 −10 −20 −30 −40 −50 −5 −4 −3 −2 −1 0 1 2 3 4 5 y Figure 1: The graph of f (y) versus y (b) [4 points] Determine the equilibrium solutions of the given differential equation; Solution: Solving f (y) = 0, i.e., y 2 (9 − y 2 ) = 0, we get the equilirium solutions: y = 0, 3, −3. (c) [7 points] Draw the phase line for the given differential equation; Solution: The phase line diagram is y 3 0 −3 Figure 2: The phase line diagram 4 (d) [3 points] Classify the stability of each equilibrium found in Part (b); Solution: • y = 3 is asymptotically stable; • y = 0 is semi-stable; • y = −3 is unstable. (e) [10 points] Sketch several graphs of solutions in the t y-plane, making sure you have at least one graph representing each type; Solution: 10 8 6 4 y(t) 2 0 −2 −4 −6 −8 −10 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 t Figure 3: Typical solutions Note that the concavity (inflection points) of the solution curves can be determined by ∂2 y checking the sign of ∂t 2 = f (y) · f 0 (y). Therefore solving f 0 (y) = 0, i.e., 2y(9 − y 2 ) + y 2 (−2y) = 0, p p we get the inflection points for the solution curves are y = 32 2 and y = − 23 2. (f) [2 points] If y(10) = −3, what is y(0)? Solution: y(0) = −3. (g) [2 points] If y(1) = −π, what is lim y(t )? Solution: lim y(t ) = −∞. t →∞ t →∞ (h) [2 points] If y(−π) = 1, what is lim y(t )? Solution: lim y(t ) = 3. t →∞ t →∞ 5 Question 4. [12 points] Solve the initial value problem x dy + (x + 2)y = cos(x)e −x , with y(π) = 0. dx On what interval around x = π is the solution defined? Solution: Divid the above equation by x, we have cos(x)e −x dy x +2 + y= dx x x This is a 1st-order linear equation with p(x) = 1 + x2 and q(x) = factor is µ(x) = e R p(x)d x =e R (1+ x2 )d x cos(x)e −x . x The integrating = e x+2 ln x = x 2 e x . Multiplying through by this function, we obtain d 2 x cos(x)e −x 2 x (x e y) = · x e = x cos(x). dx x To integrate the right side, we need to use integration by parts, Z Z Z x cos xd x = xd (sin x) = x sin x − sin xd x = x sin x − (− cos x) + c. x x Therefore, x 2 e x y = −x cos x + sin x + c ⇒ y = e −x ( sin − cos + xc2 ). x x2 This gives the general solution of the DE. To determine the constant c, we impose the initial π c π −π − cos (0 + π1 + πc2 ) ⇒ c = −π. condition y(π) = 0 ⇒ 0 = e −π ( sin π + π2 ) ⇒ 0 = e π2 x π x Thus, the solution of this IVP is y(x) = e −x ( sin − cos x − x 2 ). Note that the solution grows x2 unbounded as x approaches 0. So, the interval around x = π over which the solution is welldefined (or valid) is x ∈ (0, ∞). 6 Question 5. [10 points] Solve the following differential equation by separation of variables ex y Solution: dy ex y dx dy = e −y + e −2x−y dx = e −y (1 + e −2x ) ye y d y = 1+e −2x ex d x = (e −x + e −3x )d x R Then Z ye y d y = y y R (e −x + e −3x )d x Z ye d y = ye − and =⇒ i.e., Z e y d y = ye y − e y 1 (e −x + e −3x )d x = −e −x − e −3x + c 3 1 ye y − e y = −e −x − e −3x + c 3 1 (y − 1)e y = −e −x − e −3x + c 3 7 Question 6. [16 points] Consider the following differential equation y dy + 2e x + (ln x − 2) =0 x dx (a)[6 points] Verify the above equation is an exact equation; Solution: dy y y + 2e x + (ln x − 2) = 0 =⇒ ( + 2e x )d x + (ln x − 2)d y = 0 x dx x =⇒ y M (x, y) = + 2e x , N (x, y) = ln x − 2 x =⇒ ∂M 1 ∂N 1 = , = ∂y x ∂x x i.e., ∂M ∂N = ∂y ∂x So this DE is exact. (b)[10 points] Find the solution for x > 0 with the initial condition: y(1) = e, and determine its interval of definition. y ∂f Solution: Since the equation is exact, there exists a function f (x, y) such that ∂x = x + 2e x and R ∂f R y ∂f = ln x −2. Using the first of these equations, we have d x = ( x +2e x )d x. f (x, y) = y ln x + ∂y ∂x ∂f 2e x + g (y). Now, ∂y = ln x + g 0 (y) = ln x − 2 ⇒ g 0 (y) = −2 ⇒ g (y) = −2y. Therefore, f (x, y) = y ln x + 2e x − 2y = c is the general solution of this DE. Applying the I.C., y(1) = e, we have e ln 1 + 2e x 2e 1 − 2e = c ⇒ c = 0. We have the solution of the IVP is y ln x + 2e x − 2y = 0 ⇒ y = 2−ln , where x 2 2 2 − ln x 6= 0 ⇒ x 6= e . The interval of definition is (0, e ). 8 Question 7. [14 points] Given the following differential equation (x + ye y/x )d x − xe y/x d y = 0, (a)[4 points] Verify the above equation is homogeneous; y y ty Solution: M (x, y) = (x +ye x ), N (x, y) = −xe x . So, N (t x, t y) = −t xe t x = t N (x, y) and M (t x, t y) = ty y (t x + t ye t x ) = t (x + ye x ) = t M (x, y). Therefore, it is homogeneous. (b)[10 points] Solve the above equation with the initial condition: y(1) = 0. y y y Solution: Let y = ux, i.e., u = x . d y = ud x + xd u. So (x + ye x )d x − xe x d y = 0 ⇒ (x + uxe u )d x − xe u (ud x + xd u) =R0 ⇒ xd xR+ uxe u d x − xue u d x − x 2 e u d u = 0. Therefore, xd x = x 2 e u d u ⇒ y e u d u = x1 d x. So, e u d u = X1 d x and therefore e u = ln x + c. So e x = ln x + c ⇒ y(1) = 0 ⇒ 0 y e 1 = ln 1 + c ⇒ c = 1. So, e x = ln x + 1. 9 Question 8. [16 points] Solve the following differential equations by using an appropriate substitution: dy − (1 + t )y = t y 2 dt dy dy Solution: d t − 1+t y = y 2 ⇒ d t − (1 + 1t )y = y 2 ⇒ t n = 2, u = y 1−2 = y −1 . (a)[10 points] t dy dt = (1 + 1t )y + y 2 . du dy 1 = −y −2 = −y −2 ((1 + )y + y 2 ) dx dx t 1 −1 = −(1 + )y − 1 t 1 = −(1 + )u − 1 t R 1 so that ddut + (1 + 1t )u = −1. So p(t ) = 1 + 1t and µ(t ) = e 1+ t d t = e 6t ln t = t e t . R So ddt (t e t u) = −t e t . Also, t e t u = − t e t d t = −t e t + e t + c. u = −1 + 1t + ct e −t and y −1 = −1 + 1t + ct e −t . dy 1−x − y = dx x+y dy dy dy du ⇒ dd ux − 1 = 1−u = u1 − 1 ⇒ Solution: u = x + y, d x = 1 + d x ⇒ d x = dd ux − 1 ⇒ d x = 1−u u u ud u = d x ⇒ 12 u 2 = x + c. So, u 2 = 2x + c and (x + y)2 = 2x + c. (b)[6 points] 10 du dx = u1 . So, Question 9. [BONUS: 12 points] Find a function whose square plus the square of its derivative is 1. Solution: Let y(x) be the function we are looking for. Therefore, y(x) should be the solution of the following equation: ¡ d y ¢2 =1 y2 + dx then q ¡ d y ¢2 dy 2 =⇒ = 1 − y =⇒ = ± 1 − y2 dx dx (A) q 1 dy d y = d x =⇒ sin−1 y = x + c =⇒ y = sin(x + c). = 1 − y 2 =⇒ p 2 dx 1− y (B) q dy 1 = − 1 − y 2 =⇒ p d y = −d x =⇒ sin−1 y = −x + c =⇒ y = sin(c − x). 2 dx 1− y Therefore, the following functions can be the solution to this problem: y = sin x, y = − sin x, y = cos x, y = − cos x, y = 1, y = −1, etc. 11