Separable Differential Equation with Initial Value?
#x^2(dy/dx)=y-xy#
Initial condition : y(-1)=-1
This is what I separated it to
#(dy/y)=(1-x)/(x^(2))dx#
After integration, I have:
#ln(y)=(-xln(x)-1)/(x)#
I know that I will have to introduce "e," but I am getting tripped up here.
Any help would be appreciated.
Thanks!
Initial condition : y(-1)=-1
This is what I separated it to
After integration, I have:
I know that I will have to introduce "e," but I am getting tripped up here.
Any help would be appreciated.
Thanks!
1 Answer
# |y| = (e^(-1/x-1))/(|x|) #
Explanation:
We have:
# x^2 \ dy/dx = y-xy # with Initial condition#y(-1)=-1#
Which we can write as:
# x^2 \ dy/dx = y(1-x) => 1/y \ dy/dx = (1-x)/x^2 #
So, as indicated, the ODE is separable, so we can "separate the variables" to get:
# int \ 1/y \ dy = int \ 1/x^2 - 1/x \ dx #
Both integrals are of standard functions, so we can directly integrate, to get:
# ln |y| = -1/x - ln |x| + A #
Using the initial condition
# ln |-1| = -1/(-1) - ln |-1| + A => A = -1 #
Thus we have:
# ln |y| = -1/x - ln |x| + 1 #
Noting that we can write
# ln |y| = -1/x - lne - ln |x| #
# :. ln |y| + ln |x| + lne = -1/x #
# :. ln |xye| = -1/x #
# :. |xye| = e^(-1/x) #
# :. |y| = (e^(-1/x))/(e|x|) #
# :. |y| = (e^(-1/x-1))/(|x|) #