Solution of the problem in interlaced form

[mathmari]

Hey!

We consider the differnetial equation $$(x-xy(x))+(y(x)+x^2)y'(x)=0$$

I have found the integrating factor $\mu (x,y)=\sqrt{x^2+y^2}$.

Using this I have to find the solution of the problem in interlaced form.

Could you give me some hints what I am supposed to do? (Wondering)

I got stuck right now...

Is it maybe as follows? (Wondering) $$y'=F(y,x) \\ \Rightarrow (x-xy(x))+(y(x)+x^2)F(y,x)=0 \\ \Rightarrow y(x)(1-x)+x^2+x+x^2F(y,x)=0 \\ \Rightarrow y(x)=-\frac{x^2F(y,x)-x-x^2}{1-x}$$

 

[MarkFL]

I have never heard of "interlaced form" and I am not sure where you obtained that integrating factor. If I were going to solve this ODE, I would first write it in the form:

$$(x-xy)\,dx+(x^2+y)\,dy=0$$

We see this is not exact, so we consider:

$$M(x,y)=x-xy$$

$$N(x,y)=x^2+y$$

$$\frac{\pd{N}{x}-\pd{M}{y}}{M}=\frac{2x+x}{x(1-y)}=\frac{3}{1-y}$$

Thus, our integrating factor is:

$$\mu(y)=\exp\left(-3\int\frac{dy}{y-1}\right)=\frac{1}{(y-1)^3}$$

And so our ODE becomes:

$$-\frac{x}{(y-1)^2}\,dx+\frac{x^2+y}{(y-1)^3}\,dy=0$$

You should verify that we now have an exact equation...can you proceed?

edit: In using the above integrating factor, we should bear in mind that we lose the trivial solution $y\equiv1$. ;)

 

[MarkFL]

In order to proceed with our now exact equation, we observe that the following is true:

$$\pd{F}{x}=-\frac{x}{(y-1)^2}$$

Integrating w.r.t $x$, we obtain:

$$F(x,y)=-\frac{x^2}{2(y-1)^2}+g(y)$$

Now, to determine $g(y)$, we take the partial derivative of this w.r.t $y$, and keep in mind that $$\pd{F}{y}=\frac{x^2+y}{(y-1)^3}$$. Hence:

$$\frac{x^2+y}{(y-1)^3}=\frac{x^2}{(y-1)^3}+g'(y)$$

And so we obtain:

$$g'(y)=\frac{y}{(y-1)^3}$$

Integrating w.r.t $y$ up to a numerical constant, we get:

$$g(y)=\frac{1-2y}{2(y-1)^2}$$

And so the solution to our ODE is given implicitly by:

$$-\frac{x^2}{2(y-1)^2}+\frac{1-2y}{2(y-1)^2}=C$$

$$\frac{1-(x^2+2y)}{2(y-1)^2}=C$$