Substitution Methods and Exact Equations

[enviroengnhopeful]

There is a problem in my book which wants us to find the general solution to the given equation. I understand most of the problem it is just the integral part that is tricky. Here is the problem:

x(x+y)y' = y(x-y)

In this problem I know that you need to divide the equation by x and you get (1+y/x)y' = y/x(1-y/x). Then I know that you can substitue y/x by saying v = y/x or y = vx. Therefore, dy/dx = v + x dv/dx. Next, I know you can substitue the equation with v to make it (1+v)(v +x dv/dx) = v(1-v). From that point on, if you divide both sides by v(1+v) you can get your equation down to
x v' = v(1-v)/v(1+v)
Then I know you are supposed to put like terms on each side so it makes the equation:

v(1+v) dv/v(1-v) = dx/x.

I know the right side is ln (x), but how do you integrate the left?
Please help. Thank You (if this is incorrect in my process, please let me know as well).

 

[Fermat]

You are correct down to here:(1+v)(v +x dv/dx) = v(1-v).

Thereafter,

(v +x dv/dx) = v(1-v)/(1+v)
x dv/dx = v(1-v)/(1+v) - v = v(1-v)/(1+v) - v(1+v)/(1+v) = -2v²/(1+v)

giving,

{(1+v)/(-2v²)} dv = (1/x) dx

 

[tacman]

Your process is correct so far. To integrate the left side, you can use the substitution u = 1+v. This will give you:

∫(u-1)/(u(u-2)) du = ∫dx/x

Now, you can use partial fraction decomposition to simplify the left side:

∫(u-1)/(u(u-2)) du = ∫(1/u - 1/(u-2)) du = ∫(1/u) du - ∫(1/(u-2)) du

Using the power rule, you can integrate the first term as ln(u) and the second term as -ln(u-2). Substituting back in for u, you will get:

ln(1+v) - ln(v) - ln(1+v) + ln(v-2) = ln(x)

Simplifying, you will get:

ln(v-2) = ln(x)

Now, you can solve for v:

v-2 = x

v = x+2

Substituting back in for v, you will get:

y/x = x+2

y = x(x+2)

This is the general solution to the original differential equation.