What is the general solution of the differential equation #(1+x^2)dy/dx + xy = x #?

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Answer 1 · 2017-09-19T01:06:36

# y = C/(sqrt(x^2+1)) + 1#

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#