Differential Equations - Solved Assignments - Semester Spring 2009

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Assignment 1 Question No 1: Solve the given differential equation by Separation of variables.

e y sin 2 x dx  cos x  e2 y  y  dy  0

cos x  e 2 y  y  dy  e y sin 2 x dx  e2 y  y  sin 2 x dx   dy   y e cos x   2y e y 2sin x cos x dx  y  y  dy   e  cos x e (e 2 y  y  ye  y ) dy  2sin x dx (e y  ye  y ) dy  2sin x dx

 e dy   ye y

y

dy  2  sin x dx

    d e y   y  e  y dy     e  y dy. ( y )  dy   2cos x  c dy     e y    ye  y   e  y dy   2cos x  c   e y  ye  y   e  y dy  2cos x  c e y  ye  y  (e  y )  2cos x  c e y  ye  y  e  y  2cos x  c

is the required general solution of given D.E.

Question No 2: Solve the following Differential Equations

  x ydx   y cos  x  dy  0 y     x  y cos  x  dy   y dx y   dy y       (i ) x dx y cos  x y

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y x y cos  x y ty y f (tx, ty )    tx x ty cos  tx y cos  x ty y f ( x, y )  

As f ( x, y)  f (tx, ty) , so the given D.E. is homogeneous Let

x y

u

y

x u

d d ( x).u  x (u ) dy dx dx   2 dx u du ux dy dx  dx u2 So equation (i) becomes du dx  2

u-x u

u-x

du  dx

u-x

du  dx

u-x

du  dx

u-x

du  dx



x u

x cos u  x u x  (u 2 ) u x cos u  x u  xu x cos u  x u  xu 1  x  cos u  1  u  u 1 cos u  1 u

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du u  u dx 1 cos u  1 u 1  u  u  cos u  1 du u  -x  1 dx cos u  1 u du u  cos u  u -x  1 dx cos u  1 u du  cos u -x  dx 1 cos u  1 u du cos u x  1 dx cos u  1 u 1 cos u  1 1 u du  dx cos u x cos u  u 1 du  dx u cos u x u  1  cos u    du  dx x  u cos u u cos u  -x

1 1    sec u  du  dx x u  ln u  ln sec u  tan u  ln x  ln c u  ln cx sec u  tan u u  cx sec u  tan u x Put u  y ln

x y sec

x x  tan y y

 cx

is the required solution of given D.E.

Question No 3: Solve the following Differential Equation

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x

2

 y 2  dx  2 xy dy  0

M  x2  y 2 M  2y y

As

N  2 xy N  2y x

M  N  , so the given D.E. is exact y x

f  x 2  y 2        (i ) x f  2 xy          (ii ) x

Integrate equation (i) wrt x f ( x, y )   ( x 2  y 2 )dx x3 f ( x, y )   xy 2  h( y ) 3

Differentiate above equation wrt y f  2 xy  h / ( y ) y

Compare above equation with equation (ii) 2 xy  h / ( y )  2 xy  h/ ( y)  0  h( y )  c

So general solution of given D.E. is y

x3  xy 2  c 3

Question No 4: Solve the initial value problems

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dy  2 y  x  e3 x  e 2 x  , dx

y  0  2

dy  2 y  x (e 3 x  e 2 x ) dx  P( x)  2  3x 2x Q( x)  x(e  e ) I .F .  u ( x )  e  =e 

P ( x ) dx 2 dx

=e 2 x

y

 u ( x)Q( x)dx  c

 (e y

u ( x) 2 x

)( x)(e3 x  e 2 x ) dx  c e 2 x

 ( xe y

3x2 x

 ( xe y

3x2 x

 ( xe y

 xe 2 x  2 x )dx  c e 2 x  xe 2 x  2 x )dx  c e 2 x

x

 x)dx  c e 2 x

d x2   x  e x dx     e x dx ( x ) dx   c dx  2  y 2 x e 2 x xe x  e x   c 2 y e 2 x xe x ex x2 c y  2 x  2 x  2 x  2 x e e 2e e 1 y  xe3 x  e3 x  x 2e 2 x  ce 2 x 2

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y (0)  2 1  2  (0)e3(0)  e3(0)  (0)2 e2(0)  ce2(0) 2  2  1  c c3 So 1 y  xe3 x  e3 x  x 2e2 x  3e2 x 2

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Assignment 2 Q#1: Solve

x.

dy  y  y 2 . ln x dx

Solution dy  y  y 2 . ln x dx Dividing by x dy 1 ln x  .y  y2. dx x x It is bernoulli D.E. x.

divide by y 2 above eq dy 1 1 ln x y 2  y  dx x x Multiply by (1) above eq. dy 1 1 ln x  y 2  y   (1) dx x x put y 1  v diff w.r.t.x dy dv  y 2  dx dx

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dv 1 ln x  v dx x x It is linear in v 1

1   dx I .F  e x  e  ln x  e ln x  x 1 1 I .F  x Multiply I .F by eq (1) 1 dv 1 1 1 ln x .  ( v)   ( ) x dx x x x x 1 ln x d (v. )   ( 2 ) dx x x Taking int egrate both sides. 1 ln x  d (v. x )    ( x 2 ) dx 1 x 1 x 1 1 v.   (ln x.  . ) x 1 1 x y 1 1 x 1  .ln x  c x x 1 1 1 1  .ln x   c xy x x or 1  ln x  1  xc y

Q#2: Find the orthogonal trajectory to the family of curve Solution:

x  cy 2

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x  cy 2  (1) diff w.r.t.x dy 1  2cy dx 1 dx c put in eq (1) 2 y dy 1 dx 2 x( )y 2 y dy dy y  dx 2 x is a differential eq. of eq (1) So differential equation of family of orthogonal trajectories is \ dy 2x  dx y It can be separable. y dy  2 x dx

 y dy    2 x dx y2   x2  c/ 2 y2   2x2  c ,

where c  2c /

Q#3: If the rate of spread of dengue virus is proportional to infected & non infected persons. Then what will be number of infected persons after 2 weeks provided that a man carrying this virus, come back to his community of 10000.Further it is provided that after one week, carriers increase to 98. Solution:

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 a man carries dengue virus initially enter community  x(0)  1 {i  e no of carrier at t  0} Assuming no one leaves from community of 10,000. So, initial value problem is dy  kt 10,000  t  ; x(0)  1 dt i  e logistic equation when comparing with dP  P(a  bP) such that a  10000k , b  k dt It ' s solution is given by aP0 P(t )  ; Here P0  x(0)  1 bP0   a  bP0  e at So, 10,000k 10,000       (1) 10,000 kt k  9999e 1  9999e10,000kt  it is given that after one week , x(t ) 

Q a man carries dengue virus initially enter community \x(0)=1 {i-e no of carrier at t=0} Assuming no one leaves from community of 10,000. So,initial value problem is dy =kt 10,000-t  ; x(0)=1 dt i-e logisticequation when comparing with dP =P(a-bP) such that a=10000k, b=k dt It's solution is given by aP0 P(t)= ; Here P0 =x(0)=1 bP0 +  a-bP0  e-at So, 10,000k 10,000 = -----(1) -10,000kt k+9999e 1+9999e-10,000kt Q it is given that after one week, x(t)=

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Q#4: Determine whether the functions f1  x   e x , f 2  x   xe x and f3  x   x 2e x are linearly independent or linearly dependent.

Solution

The functions

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f 1  x   e x , f 2  x   xe x , and f 3  x   x 2 e x

ex W  e x , xe x , x 2e x   e x ex

xe x xe x  e x xe x  2e x

x 2e x x 2e x  2 xe x x 2e x  4 xe x  2e x

 2 e3 x 0

Thus the functions are linearly independent on any interval of the x-axis because for all x  R .

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Assignment 3 Question No 1: Given that y1  x 2 is a solution of x 2 y //  2 xy /  6 y  0 Find general solution of the differential equation on the interval  0,  . The DE x 2 y //  2 xy /  6 y  0 can be written as

2x 6   x 2  y //  2 y /  2 y   0 x x   2x 6  y //  2 y /  2 y  0 x x Here p( x) 

2 x

 Pdx e  y2  y1  dx ( y1 ) 2 2

  dx x 2 e  y2  x  2 2 dx (x )

 y2  x 2 

e

2

1

 xdx

x4

dx

e 2 ln x  y2  x  4 dx x 2

2

eln x  y2  x  4 dx x 2 x  y2  x 2  4 dx x 2

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 y2  x 2  x 6 dx  x 6 1   y2  x 2    6  1   x 5   y2  x 2    5   y2  

1 5 x3

General solution is given by formula y  c1 y1  c2 y2  1  y  c1 ( x 2 )  c2   3   5x  c y  c1 x 2  23 5x

Question No 2: Solve y //  4 y /  4 y  e2 x

For complementary solution y //  4 y /  4 y  0

Put y  e mx

 y /  memx  y //  m2emx So

m2emx  4memx  4emx  0  (m2  4m  4)emx  0 As emx  0 So auxiliary equation is

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m 2  4m  4  0  (m  2) 2  0  m  2, 2

So yc  c1e 2 x  c2 xe 2 x is our complementary solution For particular solution g ( x)  e2 x

So

y p  Ae 2 x  y p /  2 Ae 2 x  y p //  4 Ae 2 x So

y p //  4 y p /  4 y p  e 2 x  4 Ae2 x  4(2 Ae2 x )  4( Ae2 x )  e2 x  4 Ae2 x  8 Ae2 x  4 Ae2 x  e 2 x  0  e2 x So our assumption is wrong, we remove duplication now, as Ae2x is present in yc Now we find particular solution of form y p  Axe2x  y p /  Ae 2 x  2 Axe 2 x  y p //  4 Ae 2 x  4 Axe 2 x

Substituting in given DE y p //  4 y p /  4 y p  e2 x , we get 4 Ae 2 x  4 Axe 2 x  4  Ae 2 x  2 Axe 2 x   4  Axe 2 x   e 2 x 4 Ae 2 x  4 Axe 2 x  4 Ae 2 x  8 Axe 2 x  4 Axe 2 x  e 2 x 0  e2 x

This has again resulted in duplication

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Therefore, now we find the particular solution of the form y p  Ax2e2x y p /  2 Axe 2 x  2 Ax 2e 2 x y p //  2 Ae 2 x  8 Axe 2 x  4 Ax 2e 2 x

Substituting in given DE y p //  4 y p /  4 y p  e2 x , we get 2 Ae 2 x  8 Axe 2 x  4 Ax 2e 2 x  4  2 Axe 2 x  2 Ax 2e 2 x   4  Ax 2e 2 x   e 2 x 2 Ae 2 x  8 Axe 2 x  4 Ax 2e 2 x  8 Axe 2 x  8 Ax 2e 2 x  4 Ax 2e 2 x  e 2 x 2 Ae 2 x  e 2 x  2A  1  A

1 2

Therefore our particular solution is y p 

1 2 2x xe 2

y  yc  y p

y  c1e2 x  c2 xe2 x 

1 2 2x x e is the required general solution of given DE 2

Question No 3: Solve  D3  D  y  3  4e x , y(0)  4 , y / (0)  5 , y // (0)  1 by undermined coefficients using annihilator operator approach. For complementary solution Auxiliary equation is m3  m  0  m( m 2  1)  0  m  0, i, i

Therefore yc  c1e(0)( x )  c2e(0)( x ) cos x  c3e(0)( x ) sin x

yc  c1  c2 cos x  c3 sin x is our complementary solution

For particular solution

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Now we find the annihilator operator for function g ( x)  3  4e x D (3) = 0 And annihilator operator for 4e x Here   1 and n  1 Therefore annihilator operator for 4e x will be  D    , which will be equal to  D  1  D  1 n

Therefore D( D  1) is the annihilator operator for function g ( x)  3  4e x Now we apply D( D  1) to both sides of given DE

D( D  1)( D3  D) y  D( D  1)(3  4e x )  D( D  1)( D3  D) y  0 Auxiliary equation is m(m  1)(m3  m)  0  m(m  1)(m 2  1)m  0  m  0,0,1, i, i

 y  c1e(0)( x )  c2 xe(0)( x )  c3e x  c4 cos x  c5 sin x  y  c1  c2 x  c3e x  c4 cos x  c5 sin x The terms c1  c4 cos x  c5 sin x are present in yc Therefore remaining terms are c2 x  c3e x So y p  Ax  B  Cex

y p /  A  Ce x y p //  Ce x y p ///  Ce x Putting it in DE y p ///  y p /  3  4ex

1

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 Ce x  A  Ce x  3  4e x  A  2Ce x  3  4e x Comparing coefficients, we get

A3 2C  4 C 2

Therefore, our particular solution conforms to y p  3x  2e x

y  yc  y p y  c1  c2 cos x  c3 sin x  3x  2e x y(0)  4

 4  c1  c2 cos 0  c3 sin 0  3(0)  2e(0)  4  c1  c2  2  c1  c2  2        (i ) y /  c2 sin x  c3 cos x  3  2e x

y / (0)  5

 5  c2 sin 0  c3 cos 0  3  2e(0)  5  c3  3  2  c3  0 y //  c2 cos x  c3 sin x  2e x

y // (0)  1

1  c2 cos 0  c3 sin 0  2e(0)  1  c2  2  c2  1

Put c2  1 in equation (i) to get the value of c1

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c1  c2  2  c1  1  2  c1  1 Therefore our solution is y  1  cos x  3x  2e x

Question No 4: Solve the DE y ///  y /  tan x by using variation of parameters Given information:

u1/  tan x , u2/   sin x , u3  sin x  ln sec x  tan x For complementary solution y ///  y /  0

Put y  e mx  y /  me mx  y //  m 2e mx  y ///  m3e mx

Therefore

m3emx  memx  0 emx (m3  m)  0 As emx  0 , therefore auxiliary equation is ( m 3  m)  0  m( m 2  1)  0  m  0, i, i

So

yc  c1e(0)( x )  c2 cos x  c3 sin x yc  c1  c2 cos x  c3 sin x

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Particular Solution From complementary solution, we see that y1  1 , y2  cos , y3  sin x u1/  tan x  u1   tan xdx  u1   ln cos x

u2 /   sin x  u2    sin xdx  u2  cos x

u3  sin x  ln sec x  tan x y p  u1 y1  u2 y2  u3 y3 y p   ln cos x (1)  cos x(cos x)   sin x  ln sec x  tan x  sin x y p   ln cos x  cos 2 x  sin 2 x  sin x ln sec x  tan x As cos2 x  sin 2 x  1 Therefore

y p   ln cos x  1  sin x ln sec x  tan x y  yc  y p

y  c1  c2 cos x  c3 sin x  ln cos x  1  sin x ln sec x  tan x

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Assignment 4 Question No 1: Solve the initial value problem; also find amplitude and period of oscillation?

d 2x  16 x  0 dt 2 x(0)  10 , x / (0)  0 Put x  emt

dx  me mt dt d 2x  m 2e mt 2 dt

So the equation becomes

m2emt  16emt  0  emt  m2  16   0 As emt  0 So, the auxiliary equation becomes m2  16  0  m  4i  x(t )  c1 cos 4t  c2 sin 4t

x(0)  10

 10  c1 cos  4  0   c2 sin  4  0   c1  10 x / (t )  4c1 sin 4t  4c2 cos 4t

x / (0)  0

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 0  4c1 sin  4  0   4c2 cos  4  0   c2  0 Therefore x(t )  10cos 4t

Amplitude A  c12  c2 2  (10) 2  (0) 2  10 Period of oscillation  Here   4 Therefore Period of oscillation  

2



2 4



2

Question No 2: Find the charge q(t) on the capacitor in an LRC series circuit when L=1/4 Henry, R=10 Ohms, C=0.001 farad , E(t)=0 volts , q(0)=q0 coulombs & I(0)=0 amperes. Also, check that it is a transient or steady state solution. d 2q dq q L 2 R   E (t ) dt dt c

Putting values, we get

1 d 2q dq q  10  0 2 4 dt dt 0.001  1  d 2q dq   2  40  4000q   0 4  dt dt  d 2q dq  2  40  4000q  0 dt dt Put q  emt , q /  memt , q //  m 2e mt

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 m 2e mt  40me mt  4000e mt  0  e mt  m 2  40m  4000   0

As emt  0 , Therefore auxiliary equation is m2  40m  4000  0 Here

a=1

,

b = 40

,

c = 4000

40  (40) 2  4(1)(4000) 2(1) 40  120i m 2(1) m  20  60i m

 q(t )  c1e20t cos60t  c2e20t sin 60t q(0)  q0  q0  c1e200 cos  60  0   c2e200 sin  60  0   c1  q0  q / (t )  20c1e20t cos60t  60c1e20t sin 60t  20c2e20t sin 60t  60c2e20t cos60t q / (0)  0  0  20c1e 200 cos  60  0   60c1e 200 sin  60  0   20c2e 200 sin  60  0   60c2e 200 cos  60  0   0  20c1  60c2 Put c1 =q 0  0  20q0  60c2  60c2  20q0  c2 

q0 3

Therefore

q0 20t e sin 60t 3 1    q(t )  q0e 20t  cos 60t  sin 60t  3    q(t )  q0e 20t cos 60t 

24 | P a g e

Amplitude A  c12  c2 2

sin  

q   ( q0 )   0   3

2

2

 q0 2 

2

q0 9



10 2 q0 9



10 q0 3

=

q0 A q0

10 q0 3 3 = 10

1 q0 cos   3 A 1 q0 3 = 10 q0 3 1 = 10

sin  cos  3  10 1 10 3

tan  

   tan -1 (3)    1.249 radians

Therefore q(t ) 

10 q0e 20t  sin 60t  1.249  3

Two conditions should be fulfilled for transient solution. 1. R  0 2. lim x(t )  0 t 

Here R is 10, so first condition is true, and as there is no steady term in the solution, so lim x(t )  0 t 

Therefore, this is a transient solution

Q#3: Solve x Sol:

d 4y d 3y  6 0 dx 4 dx 3

25 | P a g e

4 3 d4y d3y 4 d y 3 d y  6  0  x  6 x 0 dx 4 dx3 dx 4 dx 3 Put x  et  t  ln x

x

Its auxiliary equation becomes (  1)(  2)(  3)  6(   1)(   2)  0   0,1, 2, 3 so we get y  c1  c2 et  c3e 2t  c4e 3t but t  ln x so y  c1  c2 x  c3 x 2  c2c3 x 3 ( x  1) n  n 2n n 0 

Q#4: Find the radius and the interval of convergence of the given power series Sol:

( x  1) n  n 2n n 0 

( x  1) n ( x  1) n 1 Here an   an 1  n 2n (n  1)2n 1 an 1 ( x  1) n 1 n 2n x 1 x 1  lim  lim  n  1 n n  a n  ( n  1)2 n  1 ( x  1) 2 n (1  )2 n The above power series converges for x  1  2 R  lim

i.e.  2  x  1  2  3  x  1 or x  (3,1) It is the required int erval of convergence.

26 | P a g e

Assignment 5 // / Q#1: Solve the differential equation 2 y  y  0 by using the power series



n solution y   cn x n 0

Solution:

Given 2 y //  y  0 

Let y   cn x n Then n 0



 2n(n  1)c x n2



  cn x n  0

n2

n

n 0

put n  2  k we get 

 2(k  2)(k  1)c k 0



k k k  2 x   ck x  0 k 0

ck ck  2  ; k  0,1, 2,3,... 2(k  2)(k  1) c c k  0; c2   0 k  1; c3   1 2  2! 2  3! c c3 c2 c k  2; c4    0 k  3; c5    1 2  4  3 2  4! 2  4  5 2  4! 2 3 4 y ( x)  c0  c1 x  c2 x  c3 x  c4 x  ... c0 2 c c c x  1 x 3  0 x 4  1 x 5  ... 2  2! 2  3! 2  4! 2  4! 2 4 3 x x x x5 y ( x)  c0 [1    ...]  c1[ x    ...] 2  2! 2  4! 2  3! 2  4! y ( x)  c0  c1 x 

Q#2: Solve the Bessel function in terms of sin x and cos x of the J 7 ( x ) 2

27 | P a g e

Solution: C onsi der t hat J v 1  x   J v 1  x  

2v Jv  x  x

As f or t a ki n g v 

5 2

J 5 1  x   J 5 1  x   2 2 J3

2

 x   J7  x   2

 2 J

2 5 x

5

2

 x

5 J5  x  x 2

5 J 7  x   J 5  x   J 3  x     (1) x 2 2 2 w e know J5

2

 x 

3 si n x 2 ( .  x x x

2 cos x)  x

2 si n x x

si n x 2 2 .  cos x x x x T hen equat i on (1) becomes J 3 / 2 ( x) 

J7

2

 x 

 5 2  3 [  2  1  si n x  x x  x 

2 3 .( cos x)]  x x

2  si n x   cos x   x  x 

Q#3: Check that the Legendre polynomials Pn ( x )  x 2  3 x  3 and Pm ( x)  2 x 3  x 2  5 x  5 are orthogonal on [0, 2].

Solution:

28 | P a g e



2



2



2



2

0

2

Pn ( x).Pm ( x)dx   ( x 2  3x  3)(2 x 3  x 2  5 x  5)dx 0

0

2

Pn ( x).Pm ( x)dx   (15  30 x  17 x 2  8x 3  5 x 4  2 x 5 )dx 0

0

0

Pn ( x).Pm ( x)dx  15 x  15 x 2 

17 x3 x6  2 x 4  x5  3 3

Pn ( x).Pm ( x)dx  6  0  6  0

Hence these functions are not orthogonal.

29 | P a g e

Assignment 6 Q#1:

Find out all the complementary solution (not particular) from the system of differential dx dy dx dy 2  5x   et ; x  5et dt dt dt dt equation

Solution: We can write them in the form of

(2 D  5) x  Dy  et ; ( D  1) x  Dy  et

And determinants are 2D  5

D

D 1

D

2D  5

D

D 1

D

 D2  4D x

et

D

5et

D

&

2D  5

D

D 1

D

y

2D  5

et

D 1

5et

we get ( D 2  4 D) x  4et ...(1) & ( D 2  4 D) y  15et ...(2) Now we find the complementary function for the two equations m 2  4m  0  m  0, 4  xc  c1  c2 e 4t & yc  c3  c4e 4t

Q#2: Find the Eigen values and Eigen vector of the given matirx

30 | P a g e

 5 1  A  0 5  5 1 

0  9 0 

Solution:

5   -1 0 | A   I | 0 -5- 9  0 5

-1

-

16   3  0

 (16   2 )  0   0, 4, 4 Here eigen values are 0,4,-4.

For

 =0, we have 5 -1 0   x  0 -5 9   y   0    5 -1 0   z  5x  y  0 5 y  9 z  0 Let y=k Then x=k/5 and z=5k/9 So vector becomes x  k / 5  1/ 5  9   y   k   k 1   k  45          z  5k / 9  5 / 9   25 So eigen vector is {9,45,25}

31 | P a g e

And eigenvectors for

 =4,-4 are {1,9,1} , {1,1,1}.

Q#3: dx dy  2y ,  8x dt dt

Find the general solution of the given system

Solution:

dx  2y , dt  dx   dt  0    dy  8  dt 

dy  8x dt 2  x  0 2 x where A   &X       0   y 8 0   y

Then det( A   I )  0 

 8

2 

 0    4, 4

32 | P a g e

for   4 , ( A   I ) X  0  4 x1  2 x2  0 &8 x1  4 x2  0 x2 k let x2  k x1   2 2  k  1   2  & eigen vector is K    2  1     k    1  for   4 , ( A   I ) X  0   4 x1  2 x2  0 &8 x1  4 x2  0 we get x1  

x2 k let x2  k x1  2 2  1 eigen vector is K 2   2     2 we get x1 

 1  1  ThereforeX1  2 e4t , X2  2 e4t     1   2  1 4t  e Hencegeneralsolutionof thesystemisthe followingX c1X1 c2X2 c1  2  4t  e So x(t) 

4t ce c2e4t 4t 1 & y(t)  ce 2c2e4t 1 2

  c   2   

1 4t e 2 2e4t

4t c2e4t   ce 1      4t 2 4t  2c2e   ce 1

    