We are given the equation,


And we have to get the dimension of solution space in $(-1,1)$.

My Attempt: I tried substituting $$x=e^z$$ and I get from that, the following,


and concluded that dimension is $2$. But, I realize that at $x=0$, my ODE in standard form will face issues, and this solution may not work. How do I go about solving this?


This will depend on precisely what you mean by a "solution" when the differential equation is singular. Every solution is of the form $y = c_1 x^2 + c_2 x^3$ on $(0,1)$ and on $(-1,0)$, but do the coefficients $c_1$ and $c_2$ for $x > 0$ and for $x < 0$ have to be the same? It is reasonable to require $y$ to be twice differentiable, since the differential equation involves $y''$. Now the second derivative at $x=0$ is $2 c_1$, so $c_1$ needs to be the same on both sides, but there's no reason for $c_2$ to be the same. Thus we have solutions of the form

$$ y(x) = \cases {c_1 x^2 + c_2 x^3 & for $x \le 0$\cr c_1 x^2 + c_3 x^3 & for $x > 0$\cr} $$ forming a vector space of dimension $3$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.