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...

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 ?(?, ?)
espectively, 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 a
itrary 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 co
esponding
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?
2
and ?2 = 3?
?
Rewriting our particular solution as...