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:

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

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