Study of a Cauchy Problem.

Question

I'm in trouble with this problem. The equation is:

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

With the initial condition $y(0)=a$

With "a" a real number.

I have to discuss local existence and uniqueness of solutions and find explicitly the solution, depending on a.

I think that it would be easy to prove local existance of solutions. I really don't know how I have to start to find explicitely the solutions. I've tried by separation of variables with no good results. Could someone give me a hint to start? Thank you very Much :).

Answer 1

It's a Bernouilli's equation

$$y'(x)=\frac{y(x)}{x-1}+x(y(x))^2$$ Divide by $y^2 \ne 0$ $$\frac {y'}{y^2}-\frac 1{y(x-1)}=x$$ Substitute $z=\frac 1 {y}$ $$z'+\frac z {x-1}=-x$$

It's simple to solve now... $$z'x-z'+z=-x(x-1)$$ $$(zx)'-z'=-x(x-1)$$ Just integrate $$z(x-1)=-\int x(x-1)dx$$ $$z=-\frac 1 {(x-1)} \int x^2-xdx$$ $$\frac 1 y=-\frac 1 {(x-1)} ( \frac {x^3}3-\frac {x^2}2 +K)$$ $$ y(x)=\frac {(1-x)} {( \frac {x^3}3-\frac {x^2}2 +K)}$$ $$ \boxed {y(x)=\frac {6(1-x)} {( {2x^3}- {3x^2} +K)}}$$

I let you finish. find the constante K with the cauchy condition..$y(0)=a$

Answer 2

For the domain of $y(x)$ I have to study

$2x^{3}-3x^{2}+\frac{6}{a}=0$ with $a\neq0$.

If $a=0$ , $y(x)=0$ is a solution.

For finding the other discontinuity points I have two idea: one is to try to scompose by Ruffini theorem the polinomia of grade 3 to obtain 3 solutions, the other is considering

$y(x,a)=2x^{3}-3x^{2}+\frac{6}{a}$

and computing the $\nabla{f}(x,a)=0$ to find and eventually classify the discontinuity point.

I want to obtain the values of $a$ such that $y(x)$ is defined on $(-\infty,1)$.

I think that "defined" means that for a certain values of $a$, $y(x)$ mustn't have discontinuity points on $(-\infty,1)$.

Thank you very much!!. .