(1 + 2y)y = x. ( x. The right-hand side is a standard integral, so in the end we have the implicit solution. y(x) + y 2 (x) = x2 2 +C.

Transcription

1 Midterm 1 33B October 1 Find the exact solution of the initial value problem. Indicate the interval of existence. y = x, y( 1) = y Solution. We observe that the equation is separable, and separate the variables: Integrate both sides in x: (1 + y)y = x. (1 + y(x))y (x) dx = x dx +C. The left-hand side can be handled by substituting z = y(x), dz = y (x)dx: (1 + z)dz = z + z = y(x) + y (x). The right-hand side is a standard integral, so in the end we have the implicit solution y(x) + y (x) = x +C. Since we must find the explicit solution, we rewrite this as ( x y ) (x) + y(x) +C = 0. This is a quadratic polynomial in y(x), so we solve using the quadratic formula: ( ) 1 ± 1 + x +C y(x) =. We now use the initial condition y( 1) = 0 to find or equivalently 0 = C ±, 1 = ± 3 + C. This forces the sign to be positive, and we obtain C = 1. Therefore the solution is y(x) = ( x ) = x 1 : that is, y(x) = x 1. The interval of existence for this solution is the largest interval containing the initial condition x = 1 for which the solution is defined. Since x 1 is not defined for 1 < x < 1, it follows that the interval of existence is (, 1 ]. 1

2 Find the solution of the initial value problem. (1 +t )y + y = (1 +t ), y(1) = 0. Since this is a linear equation, we can solve this using either integrating factors or variation of parameters. Solution 1. We divide the equation through by (1 +t ) to put it in the form y + 1 +t y = (1 +t ) 3. Then the integrating factor is (making the substitution u = 1 +t, du = tdt) e 1+t dt = e du u = e log(1+t ) = (1 +t ). Multiplying through by the integrating factor, we have Integrating in t, we find that Therefore ((1 +t ) y) = (1 +t ) 1. (1 +t ) y = arctant +C. y(t) = Using the initial condition y(1) = 0, we have arctant +C (1 +t ). Therefore C = π 0 = arctan1 +C and the solution to the IVP is = π +C. y(t) = arctant π (1 +t ). Solution. We now show how to use the method of variation of parameters. We first solve the homogeneous equation (1 +t )y h + y h = 0. This we can do by separating variables: Integrating and solving for y h, we find that y h y h (t) = e = y h 1 +t. 1+t dt = (1 +t ).

3 Now suppose that the solution to the inhomogeneous equation is of the form y + y 1 +t = (1 +t ) 3 y = vy h. Then the method of variation of parameters tells us that v = f y h = (1 +t ) 3 (1 +t ) = 1 1 +t. where f is the inhomogeneous term f = (1 +t ) 3. Therefore v(t) = and so the general solution is given by 1 dt = arctant +C, 1 +t y(t) = v(t)y h (t) = arctant +C (1 +t ). Similar calculations as in solution 1 now show that C = π and therefore y(t) = arctant π (1 +t ). 3

4 3 Use the integrating factor µ(x,y) = xy 1 to make the following equation into an exact equation. Then find the general solution. (x y 1)ydx + (1 + x y )xdy = 0. Solution. Multiplying through by the integration factor gives us the equation (xy 1 x )dx + (1 y + x y)dy = 0. Setting P(x,y) = xy 1 x and Q(x,y) = 1 y + x y, we can verify that P y so this new equation is exact. Using the relations x = xy = Q x, = P, y = Q, we solve for F. There are two equivalent ways to do this: (a) We integrate P in x to find F(x,y) = Differentiating in y, we have Since / y = Q, we see that P(x,y) dx + φ(y) = x y logx + φ(y). y = x y + φ (y). x y + φ (y) = x y + 1 y, or φ (y) = 1 y. Therefore φ(y) = logy, and the general solution is given implicitly by F(x,y) = x y logx + logy = C, where C is an arbitrary constant. (b) We integrate Q in y to find F(x,y) = Q(x,y) dy + ψ(x) = x y + logy + ψ(x).

5 Differentiating in x, we have Since / x = P, we see that or x = xy + ψ (x). xy + ψ (x) = xy 1 x, ψ (x) = 1 x. Therefore and the general solution is given by ψ(x) = logx, F(x,y) = x y + logy logx = C, where C is an arbitrary constant. 5

6 Suppose that x is a solution to the initial value problem x = x t + t, x(0) = 1. Show that x(t) > t for all t for which x is defined. There is an easy way to solve this problem, and a harder way. Solution 1 (easy). First, we show that the above equation satisfies the conditions of the uniqueness theorem for initial value problems for first-order ODEs. We set f (t,x) = x t + t. Then f is continuous everywhere in the tx-plane, and f x = 1 is also continuous in the tx-plane, so the conditions of the uniqueness theorem are satisfied. Therefore solutions to the IVP for x = x t +t are unique. In particular, two distinct solutions to the ODE x = x t + t may never cross in the tx-plane. Now we observe that x 1 (t) = t solves the IVP For x 1 (0) = 0 = 0, and x 1 = x 1 t + t, x 1 (0) = 0. x 1(t) = t = t t + t = x 1 (t) t + t. So x 1 and x are both solutions to y = f (t,y). Since x 1 (0) = 0 < 1 = x(0), the two solutions are distinct. By uniqueness, the solutions curves may never cross in the tx-plane; it follows then that x 1 (t) < x(t) for all t. That is to say, t < x(t) for all t, as claimed. Solution (harder). Since the equation is linear one can solve it using any of the valid methods for first-order linear ODEs. With some work one can show, after some careful calculations involving multiple integrations by parts, that the solution to the IVP is given by x(t) = t + e t. Since e t > 0 for all t, it follows that x(t) > t for all t. (While the description of this solution may seem simpler, it is harder in the sense that it is far easier to make a mistake while trying to solve the equation explicitly.) 6

7 5 Find the general solution for the following differential equation. y + y + y = 0. Solution. This is a second-order constant-coefficient linear equation. The characteristic equation for this equation is λ + λ + 1 = 0. Using the quadratic formula, we see that λ = ± = 1. Since the characteristic equation has repeated roots, a fundamental system for the differential equation is given by y 1 (t) = e 1 t, y (t) = te 1 t. Therefore the general solution is given by a linear combination of these fundamental solutions, that is, y(t) = Ae 1 t + Bte 1 t, where A and B are arbitrary real constants. 7