1. [20 pts] Find an integrating factor and solve the equation y 3y = e 2t. Then solve the initial value problem y 3y = e 2t, y(0) = 3.

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1 22M:034 Engineer Math IV: Differential Equations Midterm Exam 1 October 2, 2013 Name Section number 1. [20 pts] Find an integrating factor and solve the equation 3 = e 2t. Then solve the initial value problem 3 = e 2t, (0) = 3. Solution: We can use the integrating factor µ(t) = e 3t to convert the original equation into e 3t 3e 3t = e 3t e 2t or (e 3t ) = e t. Then, b integrating both sides of this equation and solving for we obtain the general solution of 3 = e 2t, namel = Ce 3t e 2t where C is an arbitrar constant. To satisf the condition (0) = 3, we must choose C = 4. Thus the unique solution of the initial value problem is = 4e 3t e 2t.

2 2. [20 pts] Solve the initial value problem dx = 2x 2 + 1, (2) = 0 and detere the interval in which the solution exists. Solution: The differential equation can be written as (2 + 1) = 2xdx. Integrating the left side with respect to and the right side with respect to x gives 2 + = x 2 + C, where C is an arbitrar constant. To satisf (2) = 0 we must have (2) 2 + (2) = C, C = 4. Hence the solution of the initial value problem is given implicitl b 2 + = x 2 4. To obtain the solution explicitl, we solve the above equation for in terms of x and we obtain = ± 1 2 4x This gives two solutions of the original differential equation, onl one of which, however, satisfies the given initial condition. This is the solution corresponding to the plus sign; thus we obtain = 1 2 ( 4x ) as the solution of the initial value problem. Finall, to detere the interval in which this solution is valid, we must find the interval containing 2 in which the quantit under the radical is positive. We find that the desired interval is x > 15 2.

3 3. [20 pts] At time t = 0 a tank contains Q 0 of salt dissolved in 100 gal of water. Assume that water containing 1 of salt per gallon 4 is entering the tank at a rate of 3 gal/, and that the well-stirred solution is leaving the tank at the same rate. Find an expression for the amount of salt Q(t) in the tank at an time t. Solution: Rate of change Rate at which Rate at which of amount of salt salt is flowing in salt is flowing out in tank in in in dq dt = Q(t) gal gal gal gal The problem ields a separable first-order linear differential equation with initial condition Q (t) = Q(t), Q(0) = Q 0. The general solution of the differential equation is Q(t) = Ce 3t , where C is an arbitrar constant. To satisf the initial condition, we must choose C = Q Thus the unique solution of the initial value problem, Q(t) = (Q 0 25)e 3t , gives the expression for the amount of salt Q(t) in the tank at an time t.

4 4. [20 pts] Solve the differential equation (3x )dx + (x 3 + 2x + e ) = 0. Solution: Let M(x, ) = 3x and N(x, ) = x 3 + 2x + e. Then M = N 3x2 + 2 and x = 3x2 + 2, and since these are equal, the equation is exact. Thus there is a potential function ψ = ψ(x, ) such that ψ x (x, ) = 3x2 + 2 ψ (x, ) = x3 + 2x + e. Our goal is to solve this sstem of equations for the function ψ(x, ). The first of these equations can be integrated with respect to x (holding constant) to find ψ(x, ) = x 3 + x 2 + h(), where the function h is an arbitrar differentiable function of, plaing the role of an arbitrar constant. B differentiating the above equation with respect to and comparing the result with the second equation in the sstem, we obtain x 3 + 2x + h () = x 3 + 2x + e, h () = e, and as a solution of the above separable equation we can take h() = e (we do not require the most general one). Finall, solutions of the original differential equation, (3x )dx + (x 3 + 2x + e ) = 0, are given implicitl b ψ(x, ) = C or x 3 + x 2 + e = C, where C is an arbitrar constant.

5 5. [20 pts] Solve the equation = (2 ). dx Also find the equilibrium solutions, classif each one as asmptoticall stable or unstable, and sketch the equilibrium solutions and several other solutions in the x-plane. Solution: To solve the equation, we integrate both sides of (2 ) and we obtain: (2 ) = dx ( ) = x + C = x + C ln 1 2 ln 2 = x + C 1 ln ln 2 = 2x + 2C 1 ln 2 = 2x + C 2 (C 2 = 2C 1 ) e ln 2 = e 2x+C 2 2 = e2x e C 2 (e C 2 > 0) = Ce 2x (C R) 2 = Ce 2x (2 ) = 2Ce 2x Ce 2x + Ce 2x = 2Ce 2x (1 + Ce 2x ) = 2Ce 2x Thus, the solutions are given b = 2Ce 2x 1 + Ce 2x. (x) = 2e2x c + e. 2x (c = 1 ) C There are two equilibrium solutions to this equation = 0 unstable, = 2 stable. We sketch the equilibrium solutions and several other solutions on the next page.

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