Solutions to Assignment 2

Transcription

1 MTHE 237 Fall 2015 Solutions to Assignment 2 Problem 1 Solve the following differential equations with the given initial conditions: a) b) y y = 2te 2t, y(0) = 1 y +y = e t, y(0) = 2 Hint: Check if the equations are exact. If needed, find appropriate integrating factors. It can be verified that the equation is not exact. Hence, we need to find an integrating factor. We learned in class that, if M y N t N is only a function of t, say is equal to P(t), then (P(t)) µ(t) = e is an integrating factor. a) In this example, we obtain P(t) = 1. Hence µ(t) = e t Hence, the equation with the integrating factor becomes: d (ye t ) = 2te t Using integration by parts for the right hand side of the equation: ye t = (2te t 2e t +C) one obtains If y(0) = 1, it follows that Hence, C = 3, and y = (2te 2t 2e 2t +Ce t ) 1 = 2+C. y(t) = 2te 2t 2e 2t +3e t b) For this example, an integrating factor is µ(t) = e t. Solving the equation leads to with y(0) = 2, it follows that c = 3/2. y(t) = 1 2 et ce t, Problem 2 Solve the following differential equation: ty +(t+1)y = t, y(ln(2)) = 1,t > 0

2 The equation is not exact. Using the same argumentwe used in the previousquestion, we find the integrating factor to be µ(t) = e t Thus, (yte t ) = e t +C and It follows by integration by parts that y(t) = ( e t t+c)e t. y(t) = (t 1+Ce t )/t. Since y(ln(2)) = 1, we find that C = 2. Thus the solution is y(t) = (t 1+2e t )/t. Problem 3 Consider a lake of (water) volume V containing Q(t) amount of pollutant, evenly distributed throughout the lake with a concentration c(t) = Q(t) V. Suppose water from a river containing a concentration k of pollutant enters the lake at a rate r (in units of volume/time) per time, and water to a canal leaves the lake with the same rate (hence, the volume of the lake is a constant). Furthermore, suppose that pollutants are also added directly to the lake by humans at a constant rate P. Express the evolution of the amount of pollutant in the lake as a differential equation. If at time t = 0, the concentration of pollutant at the lake is c 0, find an expression for the concentration c(t) as a function of time. What happens to the concentration as t? The rate of change in the amount of pollutant is = (P +rk) rc, or = (P +rk) rq/v, Hence, we have +(r/v)q = (P +rk), This is in a familiar form and the solution to this is given by: Initial value satisfies: Hence As t becomes larger, the total amount converges to Q(t) = (P +rk)(v/r) +(P +rk +C)e rt/v. c 0 V = (P +rk)(v/r) +(P +rk +C), Q(t) = (P +rk)(v/r)+(c 0 V (P +rk))e rt/v. (P +rk)(v/r),

3 and the concentration converges to k + P r. Problem 4 Obtain the general solution to the differential equation dy dx = y2 +2xy x 2, x > 0,y > 0 This is an equation with homogenous coefficients. Writing y = vx, we obtain the equation: v +x dx = v2 +2v, leading to ( 1 v 1 ) = dx v +1 x. Solving this equation leads to: v ln( x ) = ln( v +1 )+c Since the domain of the integration is the area given by x > 0,y > 0, it follows that x = Ky x+y, is the general implicit solution where K is a constant. Problem 5 Obtain the general solution to the differential equation (3x 2 +y 2 )dx+2xydy = 0. Here we observe that the equation is exact, hence, there is no need for an additional integrating factor. It follows that a function F such that F x = 3x2 +y 2 and F y = 2xy is F(x,y) = x 3 +xy 2. Hence, the general solution is given by: for some constant c. x 3 +xy 2 = c, Problem 6 Show that satisfies F(x,y) = x x 0 M(t, )+ y F(x,y) = M(x,y) x

4 if F(x,y) = N(x,y), y y M(x,y) = x N(x,y), and these are continuous in the domain of the integration. Let us first consider yf(x,y). By the fundamental theorem of calculus: y F(x,y) = ( x M(t, )+ y x 0 = y y = y (G(x,y) G(x,)) ) In the above G(x,y) is the anti-derivative of N(x,y) in the y variable. We now work with y = N(x,y) (1) x F(x,y) = ( x M(t, )+ x x 0 = M(x, )+ x = M(x, )+ y y 0 y y ) y = M(x, )+ s = M(x, )+M(x,y) M(x, ) = M(x,y) (2) x M(x,s)ds (3) In the above, (2) follows from the fact that xn(x,s) is continuous (you are encouraged to show why this holds; you can usetaylor s formula with a remainder term for the term N(x+h,s) for h > 0) and (3) follows from the hypothesis that s M(x,s) = x N(x,s), x,s. (4) Problem 7 Suppose that a rocket is launched from the surface of the earth with an initial velocity v 0 = 2gR, where R is the radius of earth (assuming a spherical earth!) and g is the gravitational acceleration at the surface of the earth and the differential equation characterization the motion is given by where x is the distance from the surface of the earth. = gr2 /(R+x) 2,

5 Find an expression for the velocity v(x) in terms of the distance x from the surface of the earth. First, observe that, by the chain rule: = dx dx = gr2 /(R+x) 2 Thus, Hence, we obtain: dx v = gr2 /(R+x) 2. gr 2 /(R+x) 2 dx = v +C. It follows that gr 2 /(R+x) = v 2 /2+C. For x = 0, the initial condition is given by v(0) = 2gR, as such C = 0. Hence, it follows that v(x) = R 2g (R+x).