1. ## Differential Equation Problem

Find the solution of the differential equation that satisfies the given initial condition. xy' + y = y^2 , y(1) = -2

I've been attempting to solve it by separable equations, by making the equation xy'=y(y-1), but my answer has not been working correctly.

2. Originally Posted by MelanieC
Find the solution of the differential equation that satisfies the given initial condition. xy' + y = y^2 , y(1) = -2

I've been attempting to solve it by separable equations, by making the equation xy'=y(y-1), but my answer has not been working correctly.
...you're not done. you should get it to $\frac 1{y(y - 1)}~dy = \frac 1x~dx$

now integrate both sides and continue

3. I've done thiat already, and by using partial fractions, I come up with:
ln(y-1/y) =lnx +c
y-1/y= x + C2
1 - 1/y = x + C2
-1/y = x + C3
y = 1/(-x + C3)

Using my initial condition, I get
-2 = 1/ (-1 + C3)
Therefore, C3 = 1/2

So my solution should be

y= 1/(-x + 1/2)

However, when I tried to check my answer by substituting it back into the original given differential equation, my answer did not work out. Can anyone help me?

4. Originally Posted by MelanieC
I've done thiat already, and by using partial fractions, I come up with:
ln(y-1/y) =lnx +c
y-1/y= x + C2
No. Taking the exponential of both sides gives (y-1)/y= C2 x where C2= $e^c$

1 - 1/y = x + C2
-1/y = x + C3
y = 1/(-x + C3)

Using my initial condition, I get
-2 = 1/ (-1 + C3)
Therefore, C3 = 1/2

So my solution should be

y= 1/(-x + 1/2)

However, when I tried to check my answer by substituting it back into the original given differential equation, my answer did not work out. Can anyone help me?

5. I know that C2 = e^c, but since e^c is a constant, i just renamed it to be C2, to make it easier to work with

6. Originally Posted by MelanieC
I've done thiat already, and by using partial fractions, I come up with:
ln(y-1/y) =lnx +c
y-1/y= x + C2 ... this step is incorrect.
$\frac{y-1}{y} = e^{\ln{x}+C} = C_2 \cdot x$ , not $x + C_2$