Homework #2 Solutions

Transcription

1 MAT Spring Problems Section.:, 8,, 4, 8 Section.5:,,, 4,, 6 Extra Problem # Homework # Solutions... Sketch likely solution curves through the given slope field for dy dx = x + y...8. Sketch likely solution curves through the given slope field for dy dx = x y. Solution: Here are slope fields with some solutions for these two problems: Determine whether Theorem.. does or does not guarantee existence and/or uniqueness for the IVP y = x ln y, y() =. If a solution does exist, is it unique? Solution: We see that the DE is already in normal form, with f (x, y) = x ln y. We then examine this function and its derivative f y (x, y) = y x around x = and y =. Since both functions are defined and continuous in a region around (, ), this IVP satisfies the hypotheses of Theorem, so it is guaranteed to have a unique solution y(x) on some interval containing x =.

2 MAT Spring..4. Determine whether Theorem.. does or does not guarantee existence and/or uniqueness for the IVP y = y, y() =. If a solution does exist, is it unique? Solution: We see that the DE is already in normal form, with f (x, y) = y = y /. We then examine this function and its derivative f y (x, y) = y / around x = and y =. While f is continuous everywhere, f y is not even defined for y =. Hence, Theorem does not apply, so neither existence nor uniqueness is guaranteed for this IVP. By inspection, we can see that y = is actually a solution to this DE passing through the initial condition (, ), so we have verified existence empirically. Since the DE is a fairly simple separable equation, we could probably solve it explicitly for y in terms of x, but we are not required to for this problem...8. Verify that if k is a constant, then the function y(x) kx satisfies the differential equation xy = y. Construct a slope field and several of these straight-line solution curves. Then determine (in terms of a and b) how many different solutions the initial value problem xy = y, y(a) = b has one, none, or infinitely many. Solution: We first check that y = kx satisfies the DE. For these solutions, y = k, so xy = x(k) = kx = y, as desired. Putting the DE in normal form, y = y/x. Letting f (x, y) = y/x, we plot a slope field with slopes given by f and some solutions y = kx: Note that f (x, y) is not defined for x =. We note that f and f y (x, y) = /x are continuous for x =, so the hypotheses of Theorem are satisfied at all points in this region. Hence, there is a unique solution to this DE going through each (a, b) with a =, namely the line y = bx/a. For a =, we see that all the linear solutions intersect at (, ), so there are an infinite number of solutions for the IC (a, b) = (, ). Conversely, no lines go through (, b) with b =, so there are no solutions to the IVP for that IC.

3 MAT Spring.5.. Find the general solution to the DE y + y =. Find the particular solution satisfying the initial condition y() =. Solution: We recognize this as a linear DE with P(x) = and Q(x) =. We then multiply the DE by the integrating factor µ(x) = e P(x) dx = e x to get (e x y) = e x y + e x y = e x. Integrating, e x y = e x dx = e x + C. Isolating y, y = + Ce x, C any real number, so this is the general solution to the DE. Applying the initial condition, = y() = + Ce = + C. Thus, C =, so y = e x is the particular solution..5.. Find the general solution to the DE y y = e x. Find the particular solution satisfying the initial condition y() =. Solution: We recognize this DE as being linear with P(x) =. Then P(x) dx = x, so an integrating factor is µ(x) = e x. Multiplying the DE by this µ(x), we have (e x y) = e x e x =, and integrating yields e x y = x + C. Hence, y = xe x + Ce x is the general solution. Applying the initial condition, = y() = ()e + Ce = C, so C =. Thus, y = xe x is the particular solution for this IVP..5.. Find the general solution to the DE xy + y = x 5. Find the particular solution satisfying the initial condition y() =. Solution: Normalizing the DE, it becomes y + x y = x4. Then P(x) = x, so P(x) dx = ln x, and an integrating factor is µ(x) = e ln x = x. In fact, we choose to take µ(x) = x. Multiplying this through the normalized form of the equation, (x y) = x y + x y = x 7. Integrating, x y = x 7 dx = 4 x8 + C. Isolating y, y = 4 x5 + Cx. Applying the IC, = y() = C = 8 + C/8. Then C/8 = 7, so C = 56, and the particular solution to the IVP is y = 4 x5 56x.

4 MAT Spring.5.4. Find the general solution to the DE xy y = x. Find the particular solution satisfying the initial condition y() =. Solution: Normalizing the DE by dividing by x, we have y x y = x. Then P(x) = x, so P(x) dx = ln x, and an integrating factor is e ln x = x. Actually, we take µ(x) = x = /x as our integrating factor for convenience. Multiplying our normalized equation by this µ(x), we have ( ) x y = x y x 4 y = x x = x. Integrating, y/x = ln x + C, so y = x ln x + Cx is the general solution to the DE. Applying the initial condition, = y() = ln + C = + C = C, so C =. Thus, the particular solution to the IVP is y = x ln x + x..5.. Find the general solution to the DE y = xy + x e x. Find the particular solution satisfying the initial condition y() = 5. Solution: The DE is already given in normal form, so we move the y term to the left-hand side to get y xy = x e x. Then this DE is linear with P(x) = x, so P(x) dx = x, and an integrating factor is µ(x) = e x. Multiplying the DE on both sides by this function, (e x y) = e x y xy = x e x e x = x. Then integrating yields e x y = x dx = x + C. Isolating y gives the general solution y = x e x + Ce x. Applying the initial condition, 5 = y() = e + Ce = C, so C = 5, and y = (x + 5)e x is the particular solution to this IVP A tank initially contains 6 gallons of pure water. Brine containing lb of salt per gallon enters the tank at gal/min, and the (perfectly mixed) solution leaves the tank at gal/min; thus, the tank is empty after exactly hour. (a) Find the amount of salt in the tank after t minutes. (b) What is the maximum amount of salt ever in the tank? Solution (a): We produce a model for this system. Let x(t) denote the amount of salt in the tank at time t, t in minutes since the brine starts flowing into the tank. The flow rate in is r i = gal/min, and the flow rate out is r o = gal/min, so the volume in the tank at time t is V(t) = 6 + ( )t = 6 t, in gallons. 4

5 MAT Spring The concentration of salt in the tank (and hence of the solution flowing out of the tank) is then c o (t) = x(t)/v(t) = 6 t x(t). Since the incoming concentration is c i = lb/gal, the net rate of change of the salt is dx dt = r ic i r o c o = ()() () 6 t x(t) = 6 t x(t), which is a linear DE. Rearranging it so that x(t) and x (t) are on the left-hand side, we have x + 6 t x =. Then P(t) = 6 t, so P(t) dt = ln 6 t, with the minus sign coming from the chain rule with 6 t. Hence, an integrating factor is e ln 6 t = 6 t. We choose to take µ(t) = (6 t), without the absolute values, since we expect solutions only for t 6. Multiplying the DE by this integrating factor, we have ( ) (6 t) x = (6 t) x + (6 t) 4 x = (6 t). Integrating, (6 t) x = (6 t) dt = (6 t) + C, so x = 6 t + C(6 t). Finally, we note that there is also an initial condition: x() =, since the tank starts with no salt. Thus, = x() = 6 + C(6 ) = C, so C = /6, and the total amount of salt in the tank at time t is x(t) = 6 t 6 (6 t). Solution (b): We find the maximum value of x(t) on [, 6]. Computing x (t), x (t) = 6 ()( )(6 t) = + (6 t). At a local extremum, x (t) =, since x (t) exists everywhere on this interval. Then = + (6 t), so (6 t) =, and t = 6. Finally, x(6 ) = 6 ( ) = = 4.9 lbs. Since x() = x(6) =, this is the maximum amount of salt in the tank. 5

6 MAT Spring Extra Problem #. Consider the DE y = x + y. (a) Plot a slope field for this DE in the region x, y. Sketch solution curves through the points (, ), (, ), and (, ). Sketch at least two other solution curves of your choice. (b) Find the general solution to this DE. Find a particular solution satisfying the initial condition y() = e. Sketch it on your slope field. Solution (a): Here is the slope field, with solutions through (, ), (, ), and (, ), as well as (, ) and (, ): Solution (b): Writing the DE as y + y = x +, we see it is linear. We multiply by the integrating factor µ(x) = e x, so (e x y) = xe x + e x. Integrating and applying integration by parts for the right-hand side, e x y = (xe x e x ) + e x + C = (x + )e x + C, so y = x + + Ce x is the general solution to the DE. We solve for C with the IC y() = e = + + Ce. Then C =, so y = x + e x. In fact, y() = =, so this is the curve through (, ), which we already sketched above. 6