For an exam I was given the following question:

$$3x y'(x)-6y(x) = x^3 y(x)^2,\quad y(x_0)=y_0$$

a) Convert this non-linear differential equation to a linear differential equation.

b) Find the solution to (a) and the given differential equation

c) When will this solution be unique and give the solution.

d) For which $(x_0,y_0)$ will there be multiple solutions and give these solutions.

I could answer the first $3$ but i just can't find (d). Any help? This is my solution so far:

a) substitute: $y(x)=1/v(x)$:

$$3x y'(x)-6y(x)=x^3 y(x)^2 \Longrightarrow v'(x)+2/x\cdot v(x)=-x^2/3$$

b) $$v(x)=\frac{C}{x^2} - \frac{x^3}{15} \Longrightarrow y(x)=\frac{15x^2}{-x^5+C}$$ (with $C$ a real constant)

c) $$y'(x)=\frac{6y(x)+x^3 y(x)^2}{3x}=F(x,\, y(x))$$

$F(x,\, y(x))$ is continuous so long $x\ne 0$.

$\dfrac{\partial F}{\partial y}$ is continuous so long $x\ne 0$.

so the given differential equation has a unique solution so long $x\ne 0$

This solution is then $$ \frac{15x^2 y_0}{x^5 y_0-x_0^5 y_0-15 x_0^2} $$

  • $\begingroup$ Hi Louise, welcome to MSE. Writing your post in Latex would be great and much more attractive for the readers... And potential answer writers. $\endgroup$ – mathcounterexamples.net Aug 13 '15 at 7:54

You find all is needed. $$ y(x)=-\frac{15x^2}{x^5+C} $$ is a solution. From equation $y(0)$ must be $0$. But if $C=0$, there is trouble at $x=0$. Ok, let $$ y_1(x)=-\frac{15x^2}{x^5+1} $$ and $$ y_2(x)=-\frac{15x^2}{x^5+2}. $$ Both are solutions and $y_1(0)=y_2(0)=0$.


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