APM2616
Assignment 3
Due date: Monday, 22 July 2024
, Question 1
(1.1)
[ y' - yx cos x = 0 ]
[ y'( pi) = 1 ]
This is a first-order linear differential equation. We can solve it using the integrating
factor method.
The standard form is:
[ y' + P(x)y = Q(x) ]
Here, ( P(x) = -x cos x ) and ( Q(x) = 0 ).
The integrating factor ( mu(x) ) is:
[ mu(x) = e^{ int P(x) dx} = e^{ int -x cos x , dx} ]
Let's compute the integral:
[ int -x cos x , dx ]
Using integration by parts:
[ u = x, , dv = - cos x , dx ]
[ du = dx, , v = sin x ]
[ int -x cos x , dx = -x sin x + int sin x , dx = -x sin x - cos x + C ]
Thus, the integrating factor is:
[ mu(x) = e^{- left(x sin x + cos x right)} ]
Multiplying both sides of the original equation by the integrating factor:
[ e^{- left(x sin x + cos x right)} y' - e^{- left(x sin x + cos x right)} yx cos x = 0 ]
Assignment 3
Due date: Monday, 22 July 2024
, Question 1
(1.1)
[ y' - yx cos x = 0 ]
[ y'( pi) = 1 ]
This is a first-order linear differential equation. We can solve it using the integrating
factor method.
The standard form is:
[ y' + P(x)y = Q(x) ]
Here, ( P(x) = -x cos x ) and ( Q(x) = 0 ).
The integrating factor ( mu(x) ) is:
[ mu(x) = e^{ int P(x) dx} = e^{ int -x cos x , dx} ]
Let's compute the integral:
[ int -x cos x , dx ]
Using integration by parts:
[ u = x, , dv = - cos x , dx ]
[ du = dx, , v = sin x ]
[ int -x cos x , dx = -x sin x + int sin x , dx = -x sin x - cos x + C ]
Thus, the integrating factor is:
[ mu(x) = e^{- left(x sin x + cos x right)} ]
Multiplying both sides of the original equation by the integrating factor:
[ e^{- left(x sin x + cos x right)} y' - e^{- left(x sin x + cos x right)} yx cos x = 0 ]
