Suppose that $\phi_1$ and $\phi_2$ are linearly independent solutions of the differential equation $$2x^2y''-(x+x^2)y'+(x^2-2)y=0,$$ and $\phi_1(0)=0$. Then the smallest positive integer $n$ such that $$\lim_{x\to 0}x^n\frac{\phi_2(x)}{\phi_1(x)}=0\ \text{is}$$
Attempt: I have started by calculating Wronskian through Abel's formula according to which $$W(\phi_1, \phi_2)(x)=W(\phi_1, \phi_2)(0)e^{\int\frac{-(x+x^2)\, dx}{2x^2}} = W(\phi_1, \phi_2)(0) \sqrt{x}e^{-\frac{x}{2}}.$$ Further, since $\phi_1$ and $\phi_2$ are linearly independent, their Wronskian must be non-zero, but $W(\phi_1, \phi_2)(0)=0$. I am confused. Please tell how to proceed.
Best Answer
Since $\phi_1(x)$ is a solution, one has $$ 2x^2\phi_1''-(x+x^2)\phi_1'+(x^2-2)\phi_1=0. \tag{1}$$ Dividing (1) by $x$ gives $$ 2x\phi_1''(x)-(1+x)\phi_1'(x)+(x^2-2)\frac{\phi_1(x)}{x}=0. $$ Using $\phi_1(0)=0$ and $\lim_{x\to0}\frac{\phi_1(x)}{x}=\phi_1'(0)$ and letting $x\to0$, one obtains $$ -3\phi_1'(0)=0,\text{ or }\phi'_1(0)=0. $$ Thus $\phi_1(x)=ax^2+O(x^3)$. Define $k(x)=\frac{\phi_2(x)}{\phi_1(x)}$ and then $\phi_2(x)=k(x)\phi_1(x)$. Since $\phi_2(x)$ is a solution, one has $$ 2 x^2 \left(\phi _1(x) k''(x)+2 k'(x) \phi _1'(x)+k(x) \phi _1''(x)\right)+\left(x^2-2\right) k(x) \phi _1(x)=x (x+1) \left(\phi _1(x) k'(x)+k(x) \phi _1'(x)\right). $$ Using (1) gives $$ \phi _1(x) \left((x+1) k'(x)-2 x k''(x)\right)-4 x k'(x) \phi _1'(x)=0. $$ So $$ \frac{k''(x)}{k'(x)}=\frac{(1+x)\phi_1(x)-4x\phi_1'(x)}{2x\phi_1(x)}=\frac12(1+\frac1x-4\frac{\phi_1'(x)}{\phi_1(x)}) $$ which gives $$ \ln(k'(x))=\frac12(x+\ln x-4\ln\phi_1(x))+c. $$ So $$ k'(x)=\frac{ce^{x/2}\sqrt x}{\phi_1^2(x)}\approx C_2x^{-7/2} $$ which gives $$ k(x)\approx C_1+ C_2x^{-5/2}. $$ Thus $$ \lim_{x\to0}x^nk(x)=\lim_{x\to0}C_2x^{n-5/2}=0 $$ implies that the smallest integer is $n=3$.