DummySchool of Mathematics and Physics, UQ MATH2001/7000, Assignment 1, Summer 2020 (1) Give an...

Question

DummySchool of Mathematics and Physics, UQ MATH2001/7000, Assignment 1, Summer 2020 (1) Give an explicit solution to the initial value problem y2 + 2xy + 2x− 1 + (2xy + x2)dy dx = 0, y(1) = 0. Show all working. (2) (a) Let A,B ∈ R be arbitrary constants. Verify that the function f(x) = A(1+x)+Bex is a solution to the differential equation xy′′ − (1 + x)y′ + y = 0. (b) Find the general solution to the differential equation xy′′ − (1 + x)y′ + y = x2e2x. Show all working. (3) Consider the matrix A =  1 4 5 6 9 3 −2 1 4 −1 −1 0 −1 −2 −1 2 3 5 7 8  (a) Give a basis for the row space of A. Show all working. (b) Give a basis for the null space of A. Show all working. (4) Let p(x), q(x) be continuous functions. Consider the ODE y′′ + py′ + qy = 0. (a) From lectures, recall that the set of all solutions to this ODE gives rise to a vector space, V . Show that V is an inner product space with inner product 〈f, g〉 = f(0)g(0) + f ′(0)g′(0). (b) Show that {cosh(x), sinh(x)} is an orthonormal basis with respect to the inner product of part (a), for the inner product space of solutions to the ODE y′′ − y = 0. (5) Consider the following data points: (−1,−14), (0,−5), (1,−4), (2, 1), (3, 23). (a) Find the least squares cubic fit y = a0 + a1x+ a2x 2 + a3x 3 to the data points. (b) Use a computational plotting tool (e.g. MATLAB) to plot the data points and the fitted curve on the same axes. (6) Consider the matrices A =  1 1 11 1 1 1 1 1  , B =  1 0 −10 0 0 −1 0 1  , C =  1 −2 1−2 4 −2 1 −2 1  , and M =  1 + α 1 1− α1 1 1 1− α 1 1 + α  , α ∈ R. You are given that the matrices A,B and C satisfy AB = BA = AC = CA = BC = CB = 0, A2 = 3A, B2 = 2B, C2 = 6C. Find an invertible matrix P and a diagonal matrix D such that M = PDP−1. Show all working. Each question marked out of 3. • Mark of 0: No relevant answer submitted, or no strategy present in the submission. • Mark of 1: The submission has some relevance, but does not demonstrate deep under- standing or sound mathematical technique. • Mark of 2: Correct approach, but needs to fine-tune some aspects of the calculations. • Mark of 3: Demonstrated a good understanding of the topic and techniques involved, with well-executed calculations. Q1: Q2(a): Q3(a): Q4(a): Q5(a): Q6: Q2(b): Q3(b): Q4(b): Q5(b): Total (out of 30):

Bhoomika · Accepted Answer

School of Mathematics and Physics, UQ 
MATH2001/7000, Assignment 1, Summer 2020 

(1) Give an explicit solution to the initial value   problem 
 
   ?2 + 2?? + 2? − 1 + (2?? + ?2)
??
??
= 0, ?(1) = 0
Show all working.
SOLUTION – 
STEPS: 
(i) Firstly finding the nature of given equation – 
The given equation does not satisfy the condition for homogeneity i.e.  
?(??, ??) = ?(?, ?) 
So it is a non- homogenous differential equation of first order of the form similar to 
??
??
+ ?(?)? = ?(?)  
But the equation is nonlinear. Rewriting the given equation 
(2?? + ?2)
??
??
+ ?2 + 2?? + 2? − 1 = 0 
Now since the equation has coefficients as functions of x and y, we test the equation 
for exactness.  
                 (?2 + 2?? + 2? − 1)?? + (2?? + ?2)?? = 0                   ---------------(1) 
Assuming the coefficients of dx and dy to be denoted as  ?(?, ?) and  ?(?, ?) 
respectively, and are continuous functions, we get 
?(?, ?) =  ?2 + 2?? + 2? − 1 
and                      ?(?, ?) = 2?? + ?2
The given equation (1) is exact if and only if              
??
??
= 
??
??
 
Therefore –  
??
??
= 2? + 2? 
And  
??
??
= 2? + 2?
So equation (1) is an exact equation.
(ii) Now we need to find the explicit solution to the problem. 
Initial value is given as y(1) = 0. 
Writing the system of 2 differential equations defining the function- 
??
??
= ?(?, ?) = ?2 + 2?? + 2? − 1
??
??
= ?(?, ?) = 2?? + ?2
Integrating the first equation over x- 
?(?, ?) =  ∫?(?, ?)?? =  ∫(?2 + 2?? + 2? − 1)?? +  ∅(?) 
  Solving it, we get  ?(?, ?) = ??2 + ?2? + ?2 − ? +  ∅(?)   
  Differentiation this w.r.t. y should give expression equal to ?(?, ?) 
and thus we can find ∅(?). 
??
??
= 2?? + ?2 + ∅′(?) = 2?? + ?2 
That implies ∅′(?) = 0,   ∅(?) = ????????, let it be C  
So the general solution to the given problem is 
 ??2 + ?2? + ?2 − ? + ? = 0 
  Applying the initial value , we get C = 0 
  So the implicit solution to the given problem is 
    ??2 + ?2? + ?2 − ? = 0 
  Explicit solution –
   ?(?) =  
−?2 ± √?4−4(?)(?2−?)
2?
     or       ?(?) =  
−?±(?−2)
2

∈  
(2) (a)  Let   ?,? ∈ ?          be arbitrary constants. Verify that the function 
 ?(?) = ?(1 + ?) + ???  is a solution to the differential equation – 
??′′ − (1 + ?)?′ + ? = 0 
(b) Find the general solution to the differential equation 
??′′ − (1 + ?)?′ + ? = ?2?2? 
Show all working.
SOLUTION-
(a) This is a homogenous differential equation of second order. 
To verify that the function  ?(?) = ?(1 + ?) + ??? is its solution, we substitute 
differentiated expressions of f(x) into the given differential equation.
Therefore, we get 
?(?) = ? = ?(1 + ?) + ??? 
?′ = ? + ??? 
?′′ = ??? 
Substituting in the main equation – 
??? =  ?(???) − (1 + ?)(? + ???) + ?(1 + ?) + ??? 
       = ???? − ? − ??? − ?? − ???? + ? + ?? + ??? 
               = 0 = ??? 
Hence proved.
(b) Given non-homogenous differential equation – 
??′′ − (1 + ?)?′ + ? = ?2?2?   ⋯⋯⋯⋯⋯(1) 
 
It forms the type as follows – 
?′′ + ?(?)?′ + ?(?)? = ?(?) 
 
      Using method of Variation of Parameters, we need to evaluate 2 components to find the 
general solution –
First, the complementary solution ??(?)  – i.e. the general solution to the corresponding 
homogenous equation 
??′′ − (1 + ?)?′ + ? = 0 
First part of this question already gave the solution and we verified it. So
??(?) = ?(1 + ?) + ??
? ⋯⋯⋯⋯⋯⋯(2)
Next, the particular solution ??(?) to the given non-homogenous equation so that  
??(?) =  ??(?) + ??(?)
So now we find ??(?) for the problem – 
 We replace the constants A,B of equation(2) with unknown functions 
?1(?) ??? ?2(?)  to get to the particular solution, such that – 
??(?) = (1 + ?)?1(?) + ?
??2(?) 
= ?1?1(?) + ?2?2(?)
Where (1 + ?) = ?1 and ?
? = ?2 form the linear combination.
To find the 2 unknown functions, we need 2 conditional equations to get the solution. 
 First 
?1
′?1 + ?2
′ ?2 = 0 
?1
′ (1 + ?) + ?2
′ ?? = 0 
 Substitution 
?1
′ ?1
′ + ?1
′?2
′ =
?(?)
??????????? ?? ?′′
 
So we get,
?(?1
′ . 1 + ?1
′ . ??) = ?2?2? 
→ ?1
′ + ?2
′ ?? = ??2?
Subtracting the 2 conditional equations – 
−?. ?1
′ = ??2? 
?1
′ = −?2?   ⋯⋯⋯⋯⋯⋯(3) 
And therefore 
?2
′ = ??(1 + ?)  ⋯⋯⋯⋯⋯(4)
Integrating equation (3) and (4), we get 
?1 = −
?2?

School of Mathematics and Physics, UQ MATH2001/7000, Assignment 1, Summer 2020 (1) Give an explicit solution to the initial value problem y2 + 2xy + 2x− 1 + (2xy + x2)dy dx = 0, y(1) = 0. Show all...

Answer To: School of Mathematics and Physics, UQ MATH2001/7000, Assignment 1, Summer 2020 (1) Give an explicit...

Answer To This Question Is Available To Download

Related Questions & Answers

Submit New Assignment