For each $\lambda\in \mathbb{R}$, let $\varphi_{\lambda}$ : $J_{\lambda}\rightarrow \mathbb{R}$ denote the solution to the following initial value problem: $$ \frac{dy}{dx}=y^{2}+\lambda\sin^{2}x, y(0)=0, $$ where $J_{\lambda}$ is the maximal interval of existence.
1. Prove that for each $\lambda>0, J_{\lambda}$ exists and the above solution is unique; prove that $\varphi_{\lambda}$ is bounded.
2. Let $\Phi(\lambda, x)=\varphi_{\lambda}(x)$. Calculate $\displaystyle \frac{\partial\Phi}{\partial\lambda}(0, x)$
For the first part:
"Let $f(x,y)=y^{2}+\lambda\sin^{2}x$.
Since $\displaystyle \frac{\partial f}{\partial y}=2y$ is continuous, thus $f$ satisfies a local Lipschitz condition with respect to $y$, hence by the uniqueness theorem, the solution $\varphi_{\lambda}$ is unique.
Also by another theorem, the IVP has a unique maximal interval of existence, thus $J_{\lambda}$ exist. "
I am stuck at the next part as I not sure how to prove that $\varphi_{\lambda}$ is bounded.
Also, by using methods to solve a riccati equation:
let $P(x)=\lambda\sin^{2}x, Q(x)=0, R(x)=1.$
Then since $y_{0}=0$ is a particular solution of this IVP, I first define $z=\displaystyle \frac{1}{\varphi_{\lambda}}.$
Next, by theorem, $$ z'=-(0+2(0)(1))z-1=-1 $$ $$ z=-x+c_{1} $$ So, $\varphi_{\lambda}$=$\displaystyle \frac{1}{c_{1}-x}.$
But since $y(0)=0,$ then I will get some undefined $c_{1}.$ Thus i am confused at this part.
Since I am unable to get $\varphi_{\lambda}(x)$, I couldn't attempt part 2.