Mathematische Methoden Übungstest, Physik, Astronomie 2./3. Semester

8/14/2019 Mathematische Methoden bungstest, Physik, Astronomie 2./3. Semester

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Prufung Do. 27.01 Test A (gesamter Stoff )

1 (5 Punkte)

Seien a, b positive reelle Zahlen ( a , b R , a , b > 0). Bestimmen Sie die

komplexen Zahlen z, welche die folgende Gleichung genugen:( a )z = b.

2 (5 Punkte)

Bestimmen Sie Typ, Ordnung und Grad der Di ff erentialgleichung:

y + 4 xy ln x = e x2 (2ln x 1) ,

und l osen Sie sie

3 (5 Punkte)

Bestimmen Sie Typ, Ordnung und Grad der Di ff erentialgleichung

y + 4 y + 4 y = e 2x

x 2 ,

und l osen Sie sie

4 (5 Punkte)

Fuhren Sie das folgende reelle Integral auf ein komplexes Integral zur uckund l osen Sie es.

2

0cos4 d .

Zusatzfrage: Wie konnen Sie das Residuum einer Funktion ablesen, wennSie ihre Laurent Reihe bestimmt haben?

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5 (5 Punkte)

Uberpr ufen Sie eine mathematische Eigenschaft der Laplace-Transformation.Losen Sie das folgende Anfangswertproblem mit Hilfe der Laplace-Transformation.

y (4) (x ) + 5 y (x ) + 4 y(x ) = u (x )y(0) = y (0) = y (0) = y (0) = 0

wobei u c (x ) die Heavyside Funktion ist:

u c (x ) =0, x < c1, x c

.

6 (5 Punkte)

Bestimmen Sie die Losung des folgenden inhomogenen Systems von Di ff er-entialgleichungen, wobei a R , a = 1.

y (x ) = 2 a

1 0 y + 1 2 e

x .

Gutes Gelingen!

Schlussel:Note 19 Sehr gut ( 1)16 Note < 19 Gut ( 2)13 Note < 16 Befriedigend ( 3)10 Note < 13 Genugend ( 4)Note < 10 Nicht Gen ugend ( 5)

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Pr ufung 27.01 Test A

1

We can take the complex logarithm of both sides of the given equation,obtaining:

z Ln (a) = Ln ( b), z = Ln (b)Ln (a)

.

We have:

Ln (a) = Ln ae i ( +2 n ) = i( + 2 n ) + ln a.

Ln (b) = Ln bei ( +2 k ) = i( + 2 k ) + ln b,

and, consequently:

z = i( + 2 k ) + ln bi( + 2 n ) + ln a

.

2

This is a linear ODE. Given a linear ODE:

y + r(x)y = s(x),

we know that the solution y(x) is given by:

y(x) = eR r ( x )dx s(x)eR r ( x )dx dx + K ,therefore the solution of the given ODE can be directly written as:

y(x) = e4 R x ln xd x ex 2 (2ln x 1)+4 R x ln xd x dx + K .1


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The integral

x ln xdx can be obtained by parts, namely:

x ln xdx = x2

2 ln x x

2

2

1x

dx = 14

x2(2ln x 1),therefore we obtain:

y(x) = ex2 (2ln x1) ex 2 (2ln x 1)+ x 2 (2ln x 1)dx + K = ( x + K )ex 2 (2ln x 1) .

3

We nd rst the complementary solution, namely the solution of the corre-sponding homogeneous ODE. The characteristic equation is:

2 + 4 + 4 = 0 , ( + 2) 2 = 0 .This equation has thus a double root = 2. The two functions that buildthe fundamental set of solutions are:

y1(x) = e2x , y2(x) = xe2x ,

and the complementary solution is:yc(x) = c1y1(x) + c2y2(x) = ( c1 + c2x)e2x .

Since the inhomogeneous function g(x) = e2x /x 2 is not a combination of sines, cosines, exponential functions and polynomials, the only viable methodto nd the particular solution is the method of the variation of constants.With this method, the particular solution is given by:

y p(x) = y2(x) g(x)y1(x)W (x) dx y1(x) g(x)y2(x)W (x) dx,where W (x) is the Wronskian. In our case, the Wronskian is given by:

W (x) = e2x xe2x

2e2x (1 2x)e2x= e4x (1 2x + 2x) = e4x .

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5

We calculate the Laplace transform of both sides of the given di ff erentialequation. To do that, we must remind that:

L y(n )(x) = sn Y (s) sn 1y(0) y(n 1) (0),where Y (s) is the Laplace transform of y(x) and that:

L {uc(x)} = ecs

s .

By using the given initial conditions, we get:

s4Y (s) + 5 s2Y (s) + 4 Y (s) = e s

s , Y (s) =

e s

s (s4 + 5 s2 + 4).

Since s4 + 5 s2 + 4 = ( s2 + 4) ( s2 + 1), we can also write:

Y (s) = e s

s (s2 + 4) ( s2 + 1).

We must now decompose the fraction 1s (s 2 +4)( s 2 +1) with the method of thepartial fractions. We have:

As

+ Bs + C

s2 + 4 +

Ds + E s2 + 1

= 1

s (s2 + 4) ( s2 + 1)

A s4 + 5 s2 + 4 + ( Bs + C ) s3 + s + ( Ds + E ) s3 + 4 s = 1.

By comparing terms with like power of s, we obtain the following system of equations:

A + B + D = 0C + E = 05A + B + 4D = 0

C + 4E = 04A = 1

,

A = 14B = 112C = 0

D = 13

E = 0

.

In the end, the function Y (s) is given by:

Y (s) = e s14

1s

+ 112

ss2 + 4

13

ss2 + 1

.

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By means of the tabulated Laplace transforms (in particular, using the rela-

tion L {uc(x)f (x c)} = ecs F (s)), we obtain:y(x) =

u (x)12

{cos[2(x )] 4 cos(x ) + 3 } .

6

We have to solve rst the homogeneous system of ODEs:

y (x) = 2 a

1 0y .

The eigenvalues of this system are the solutions of the algebraic equation det(A I ) = 0. We have:

2 a

= 0 , 2 2 + a = 0, 1,2 = 1 1 a.We can distinguish now two cases:

a < 1

In this case, 1 = 1 + 1 a and 2 = 1 1 a are real, distinctroots of the characteristic equation. The eigenvector p (1) , corresponding tothe eigenvalue 1 is the solution of the equation ( A 1I )p = 0 , namely:

1 1 a a1 1 1 a

p1 p2

= 00 .

These are two linearly dependent equations. In fact, if we multiply the secondone by 1 + 1 a we obtain the rst one. It is enough to solve the rstof these two equations, namely (1 1 a) p1 + ap2 = 0. If p1 = a, then itmust be p2 = 1 a 1, namely the eigenvector p (1) is given by:

p (1) = a

1 a 1.

The eigenvector p (2) , corresponding to the eigenvalue 2 is instead the solu-tion of the equation:

1 + 1 a a1 1 + 1 a

p1 p2

= 00 .

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These equations are once again linearly dependent (if we multiply the second

one by 1 1 a we obtain the rst one). It is enough to solve the rstof these two equations, namely (1 + 1 a) p1 + ap2 = 0. If p1 = a, then itmust be p2 = 1 a 1, namely the eigenvector p (2) is given by:p (2) = a

1 a 1.

The complementary solution y c (x) is thus given by:

y c (x) = c1 a 1 a 1

e(1+ 1a )x + c2 a

1 a 1e(1 1a )x .

We can nd the particular solution y p (x) with the method of the undeter-mined coefficients. Since a = 1, the particular solution must be of the typey p (x) = bex , where the vector b must be determined. Upon substitutioninto the given system of ODEs (and cancelling out the term ex ), we obtain:

b = Ab + 1

2, (A I )b =

12 .

This corresponds to the system of equations:

b1 + ab2 = 1

b1 b2 = 2,

b1 = 1 ab21 + ( a 1)b2 = 2

,

b2 = 1a1

b2 = a +1

a

a 1 = 1

2a

a 1.

In the end, the vector b is given by:

b = 1a 1

11 2a

,

and the general solution is:

y = c1 a 1 a 1

e(1+ 1a )x + c2 a

1 a 1e(1 1a )x +

1a 1

11 2a

ex .

a > 1

In this case, 1 = 1 + i a 1 and 2 = 1 i a 1 are complex conju-gated roots of the characteristic equation. The eigenvector p (1) , correspond-ing to the eigenvalue 1 is the solution of the equation:

1i a 1 a1 1 i a 1

p1 p2

= 00 .

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These are two linearly dependent equations. In fact, if we multiply the second

one by 1 + i a 1 we obtain the rst one. It is enough to solve the rstof these two equations, namely (1 i a 1) p1 + ap2 = 0. If p1 = a, then itmust be p2 = i a 1 1, namely the eigenvector p (1) is given by:p (1) = ai a 1 1

.

We do not need to nd the second eigenvector p (2) , corresponding to theeigenvalue 2. In fact, we know from the theory that the real and imaginaryparts of p (1) e 1 x are a suitable (and real-valued!) basis of the solutions of thegiven system of ODEs. Naming s = a

1 for sake of clarity, we have:

p (1) e 1 x = ais 1e(1+ is )x = ex ais 1

[cos(xs ) + i sin(xs )]

= ex a cos(xs ) + ia sin(xs )is cos(xs) s sin(xs ) cos(xs ) i sin(xs )Consequently:

u = Re p (1) e 1 x = ex a cos(xs )

s sin(xs )

cos(xs )

v = Im p (1) e 1 x = ex a sin(xs )s cos(xs ) sin(xs )The complementary solution does not change compared to the previous case.The general solution is thus:

y = c1 a cos(xs )

s sin(xs ) cos(xs )+ c2

a sin(xs )s cos(xs ) sin(xs )

+ 1a 1

11 2a

ex .

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Mathematische Methoden Übungstest, Physik, Astronomie 2./3. Semester

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Transcript of Mathematische Methoden Übungstest, Physik, Astronomie 2./3. Semester