MATH2000 Assignment 1: ODE Solutions and IVPs - Summer 2017-2018

This assignment provides detailed solutions to several problems related to ordinary differential equations (ODEs) and initial value problems (IVPs). The first question involves finding the general solution to a given ODE using an integrating factor. The second question focuses on solving initial value problems for a second-order linear homogeneous differential equation, determining the solutions ya and yb, and then finding a general solution in terms of ya and yb. The third question requires finding the general solution to another ODE, involving division by x^2 and determining the Wronskian for a fundamental set of solutions. Finally, the fourth question deals with the inverse hyperbolic cosecant function, finding its derivative, and evaluating an integral involving it. The assignment demonstrates a strong understanding of ODEs, IVPs, and related mathematical techniques.

Q1
The general solution to the ODE
y + (2 x− y e y ) dy
dx =0
Let:M = y : ∂ M
∂ y =1 N=2 x− y e y : ∂ N
∂ x =2 since ∂ M
∂ y ≠ ∂ N
∂ x Find the integrating factor I:
I =e∫ P dxwhere P= 1
2 x − y ey I=e∫ 1
2 x− y e y dx
=eln ( |2 x− y ey
|) =2 x− y e y
Multiply the equation using I
( 2 x− y e y ) dy
dx + y=0
Integrating both sides
2 xy− ( y−1 ) e y +C
Q2
Part a
y' ' − y=0 y ( 0 ) =1: y' ( 0 )=0 m2−m=0⇒ m=1∨−1
Therefore the general solution is:
ya=c1 et +c2 e−t ya
' =c1 et −c2 e−t
Inserting the initial conditions:
1=c1 e0+ c2 e−0 ⟹ c1+c2 =10=c1 e0−c2 e−0 ⟹ c1 −c2=0∴ c1= 1
2
'
c2=1
2
Thus, the exact solution is:
ya= 1
2 ( et +e−t )
Part b
y' ' − y=0 y ( 0 ) =0 : y' ( 0 ) =1
From part a, the value of c1∧c2 is given as:
0=c1 e0 +c2 e−0 ⟹ c1 +c2=0 1=c1 e0−c2 e−0 ⟹ c1−c2=1∴ c1= 1
2
'
c2=−1
2
Thus, the exact solution is:
yb= 1
2 ( et −e−t )
Part c

The general solution to y' ' − y=0 is ya=c1 et +c2 e−t given that y= A y a+ B yb
∴ y= 1
2 ( A ( et + e−t ) + B ( et−e−t ) ) ⟹ y0= A y'= 1
2 ( A ( et −e−t )+ B ( et + e−t ) ) ⟹ y1=B
Therefore:
∴ y= 1
2 ( y0 ( et + e−t ) + y1 ( et−e−t ) )
Q3
General solution to:
x2 y' ' −x y' + y=xln (|x|)Divide by x2 :
y' ' − 1
x y' + 1
x2 y = 1
x ln (|x|)The Wronskian for the fundamental set of solutions is;
W =
[ x
6 ln2 (|x|)
x
6 ln (|x|) c1 +c2 ln (|x|) ]Therefore, the general solution is:
y= x
6 ( c1 +c2 ln (|x|) + ln3 (|x|) )
Q4
cosec h−1 ( x ) =ln ( √ 1+ 1
x2 + 1
x ) d
dx (cosec h−1 ( x ) )=−1/¿
∫ ( cosec h−1 ( x ) ) dx=xcosec h−1 ( x ) + ln
|x ( 1+ √ 1+ x2
x2 )|+ c
Thus;
cosec h−1 ( x ) =
{ln ( 1− √ 1+x2
x ) ,∧x <0
ln ( 1+ √ 1+ x2
x ) ,∧x >0