What is the general solution of the differential equation #(1+x^2)dy/dx + xy = x #?
1 Answer
# y = C/(sqrt(x^2+1)) + 1#
Explanation:
We have:
# (1+x^2)dy/dx + xy = x # ..... [A]
We can rearrange this First Order differential equation as follows:
# (1+x^2)dy/dx = x - xy #
# :. (1+x^2)dy/dx = x(1-y) #
# :. 1/(1-y) dy/dx = x/(1+x^2) #
This is now separable, so we can "seperate the variables" to get:
# int \ 1/(1-y) \ dy = int \ x/(1+x^2) \ dx #
We can now integrate to get:
# -ln| y-1 | = 1/2 ln | x^2+1 | + ln A #
# " " = ln sqrt(x^2+1) + ln A #
# " " = ln Asqrt(x^2+1) #
And so:
# ln| y-1 | = ln ( 1/(Asqrt(x^2+1)) ) #
Taking exponentials (or anti-logarithms) and noting that the logarithm argument of the RHS must be positive, we have:
# y-1 = 1/(Asqrt(x^2+1)) #
Leading to the General Solution:
# y = 1/(Asqrt(x^2+1)) + 1#
Or:
# y = C/(sqrt(x^2+1)) + 1#
