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.

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.

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.

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 4x2 15 1 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.

Find an 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.

Our goal is to solve this sstem of equations for the function ψ(x, ).

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.

Solution: To solve the equation, we integrate both sides of (2 ) and we obtain: (2 ) = dx 1 1 2 ( + 2 2 ) = x + C 1 1 2 + 1 = x + C 1 2 2 1 2 ln 1 2 ln 2 = x + C 1 ln ln 2 = 2x + 2C 1 ln 2 = 2x + C 2

Autonomous Equations / Stability of Equilibrium Solutions. y = f (y).

Autonomous Equations / Stabilit of Equilibrium Solutions First order autonomous equations, Equilibrium solutions, Stabilit, Longterm behavior of solutions, direction fields, Population dnamics and logistic