Prove for all $f\in L^2(0,1)$, there exists a unique $T(f) \in H^1(0,1)$, such that $$-T(f)''(x)+xT(f)'(x)+T(f)(x)=f(x) \quad \forall x\in(0,1).$$
I know we need to use Lax-Milgram theorem and find a bilinear form. I do not know to find the Bilinear form.
We show that there is a unique weak solution $T=T(f)$ in $H^1_0(0,1)$ in the sense that for every $U\in H^1_0(0,1)$, $$ B[T,U]=\int_0^1 f(x)U(x) dx ,$$ where $$B[T,U]=\int_0^1 T'(x)U'(x) + x T'(x) U(x) + T(x)U(x) \, dx.$$ We explicitly choose $$\|T\|_{H^1_0(0,1)} := \|T\|_{H^1(0,1)} = \sqrt{\|T\|_{L^2(0,1)}^2 + \|T'\|_{L^2(0,1)}^2 }.$$ Clearly, $B$ is a bounded bilinear operator $|B[T,U]| \lesssim \|T\|_{H^1_0(0,1)} \|U\|_{H^1_0(0,1)}$. We only check coercivity: \begin{align} \|T\|_{H^1_0(0,1)}^2 &= \|T\|_{L^2(0,1)}^2 + \|T'\|_{L^2(0,1)}^2 \\ &= B[T,T] - \int_0^1 xT'T\, dx \\ & \le B[T,T] + \int_0^1 |T'T| \,dx \\ & \le B[T,T] + \|T\|_{L^2(0,1)} \|T'\|_{L^2(0,1)} \\ &\le B[T,T] + \frac12( \|T\|_{L^2(0,1)}^2+ \|T'\|_{L^2(0,1)}^2) \\ &= B[T,T] + \frac12\|T\|^2_{H^1_0(0,1)}. \end{align} Thus $$B[T,T] \ge \frac 12 \|T\|^2_{H^1_0(0,1)},$$ so $B$ is coercive. Hence, Lax-Milgram applies, giving the existence of a unique weak solution.