Universitetet i Bergen Det matematisk-naturvitenskapelige fakultet Eksamen i emnet Mat131 - Differensiallikningar I Onsdag 25. mai 2016, kl.

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1 1 MAT131 Bokmål Universitetet i Bergen Det matematisk-naturvitenskapelige fakultet Eksamen i emnet Mat131 - Differensiallikningar I Onsdag 25. mai 2016, kl Oppgavesettet er 4 oppgaver fordelt på 2 sider. Tillatte hjelpemidler: Ingen Skriv kun EN oppgave per ark. Oppgavene skal sorteres i rekkefølge før besvarelser leveres inn. Oppgaver som ligger i feil orden risikerer å ikke bli rettet. Les nøye gjennom oppgavesettet. Alle svar skal begrunnes, men begrunnelsene skal være korte. Det må være nok mellomregning til at framgangsmåten går tydelig frem av det du skriver. Det blir gitt godt med poeng for riktig framgangsmåte, selv om du ikke kommer frem til korrekt svar. Alle deloppgaver teller likt. (1 En differensialligning er gitt ved y + 6y + 9y = 0, t 0. (1 (a Finn den allmenne løsningen til (1. Vis at det er en løsning og at den utgjør en fundamentalmengde. (b Løs ligning (1 med initialbetingelser y(0 = 0, y (0 = 1. Solution: (a Ansatz, y(t = exp(rt gives the characteristic equation: 0 = r 2 + 6r + 9 = (r + 3 2, r 1,2 = 3, double root We have one solution y(t = c 1 e 3t. To obtain a second make the following ansatz: y = c 1 te 3t and insert in the equation. To this end, compute its derivatives: Into (1: y = c 1 e 3t 3c 1 te 3t y = 3c 1 e 3t + 9c 1 te 3t 3c 1 e 3t = 6c 1 e 3t + 9c 1 te 3t 6c 1 e 3t + 9c 1 te 3t + 6(c 1 e 3t 3c 1 te 3t + 9c 1 te 3t = 0

2 2 6c 1 + 9c 1 t + 6c 1 18c 1 t + 9c 1 t = 0 The equation is satisfied for all t implying that y(t = c 1 te rt is also a solution. We show that we now have a complete set by computing the Wronskian. ( ( y1 y W (y 1, y 2 = det 2 e 3t te y 1 y 2 = det 3t 3e 3t e 3t 3te 3t = e 6t ((1 3t + 3t = e 6t which is non-zero. (b Solve with y(0 = 0, y (0 = 1 where the general solution is y(t = (c 1 t + c 2 e 3t. Hence, y(t = te 3t is the solution. y(0 = c 2 e 3t t=0 = 0 1 = y (0 = c 1 e 3t 3c 1 te 3t t=0 = c 1. (2 (a Finn løsningen, y(x, til differensialligningen y (x = 2xy2 + 1, y( 1 = 2 2x 2 y og bestem intervallet hvor løsningen er definert. (b Løs differensialligningen og bestem løsningsintervallet. ty (t + 2y(t = t 2 t + 1, y(1 = 1 2 Solution: (a Rewrite the equation: (2xy x 2 y dy dx = 0 d dx (x2 y 2 + x = 0 Hence, x 2 y 2 + x = C. Use the initial condition, Solving for y gives C = ( ( 1 = 3. y 2 = 1 x + 3 x 2.

3 3 y(x is singular at x = 0. Since the initial condition was given at x = 1, we conclude that < x < 0 is the solution interval. In this interval the righthand side is positive and we can take the square root and obtain the solution: 1 y(x = x + 3 x. 2 The positive branch satisfies the initial condition. (b Solve Divide by t: ty + 2y = t 2 t + 1, y(1 = 1 2 y + 2 t y = t t The integrating factor is: 2 µ(t = exp( t dt = exp(2 ln t = exp(ln t 2 = t 2. Multiply the equation by the integrating factor: (t 2 y = t 3 t 2 + t Integrate: Consequently, t 2 y = t 3 t 2 + t dt = 1 4 t4 1 3 t t2 + c y = 1 4 t2 1 3 t c t 2 Use the intital condition: 1 2 = c and we see that c = 1. The solution is then, 12 defined on the interval 0 < t <. y = 1 4 t2 1 3 t t 2 (3 (a Finn den allmenne løsningen til, ( ( d u(t 0 1 = dt v(t 1 0 ( u(t v(t og klassifiser likevektspunktet.

4 4 (b Finn den allmenne løsningen til ( 1 4 der A = 1 3. x = Ax (a The critical point is (0, 0 T. The eigenvalues of A are given by 0 = det(a λi = ( λ 2 1, λ = ±1. Since the matrix is symmetric we already know that there is a full set of eigenvectors and we conclude that the critical point is a saddle. For the general solution we need the eigenvectors. We solve A w = λ w for the two eigenvalues and obtain: (1, 1 T, λ = 1 (1, 1 T, λ = 1 The general solution is: w(t = c 1 ( 1 1 e t + c 2 ( 1 1 e t. (b Find the eigenvalues of A. (1 λ( 3 λ + 4 = 0 results in a single eigenvalue λ = 1 with the eigenvector v 1 = (2, 1 T. Hence, one solution is: ( 2 x 1 (t = e t. 1 We make the following ansatz for a second solution. x 2 (t = e t (v 1 t + v 2. where v 2 is an unknown vector. Insert x 2 in the equation and manipulate: (e t (v 1 t + v 2 = Ae t (v 1 t + v 2 e t (v 1 t + v 2 + e t v 1 = Ae t (v 1 t + v 2 v 2 + v 1 = Av 2 v 1 = (A + Iv 2 Solving this results in v 2 = (0, 1/2 T. ( Then x 2 = e t 2t and the general soltution is, x(t = c t + 1/2 1 x 1 (t + c 2 x 2 (t. Note that there are arbitrary constants implying that the solution may be expressed in many different ways.

5 5 (4 Løs 2 u t 2 med initial- og randbetingelser = 2 u, 0 < x < L, t > 0, x2 u(x, 0 = f(x = 5 cos( 10πx L, u t (x, 0 = 0, u x (0, t = u x (L, t = 0. Solution: Solution by separation of variables. The ansatz, u(x, t = X(xT (t, is inserted into the wave equation and results in X X = T T = λ Hence, X satisfies X + λx = 0 with solutions X(x = a cos(µx + b sin(µx where λ = µ 2. (Note that λ = 0 and X = 1 is also a solution that is included in the first form by allowing µ = 0. From the boundary conditions we deduce that X (0 = X (L = 0. X (0 = a sin(µ0 + b cos(µ0 b = 0 X (L = a sin(µl + b cos(µl µ = nπ L, n = 0, 1, 2,... We have X n (x = a n cos(µx. The equation for T is, T + λ n T = 0 and T n (t = k 1 cos( nπt L + k 2 sin( nπt L, (T n t (t = k 1 nπ L sin(nπt L + k 2 nπ L cos(nπt L. Here, T t (0 = 0 implying k 2 = 0. The fundamental solutions are We write the solution as: u n (x, t = cos( nπt L cos(nπx, n = 0, 1, 2, 3,... L u(x, t = n=0 a n cos( nπt L cos(nπx L

6 6 The coefficients are calculated using the initial data as: a n = 2 L L We get a 10 = 5 and otherwise a n = 0. Hence, Lykke til! Magnus Svärd 0 5 cos( 10πx L cos(nπx L dx u(x, t = 5 cos( 10πt L cos(10πx L.