The problem asks us to solve the Cauchy problem, which is the initial value problem for the differential equation $\frac{y'}{3x^2 - 2\sin x} + \sqrt{y} = 0$ with the initial condition $y(0) = 1$.

1. Problem Description

The problem asks us to solve the Cauchy problem, which is the initial value problem for the differential equation

$\frac{y'}{3x^2 - 2\sin x} + \sqrt{y} = 0$

with the initial condition $y(0) = 1$.

2. Solution Steps

First, rewrite the equation as

$\frac{dy}{dx} = -(3x^2 - 2\sin x)\sqrt{y}$.

Then, separate the variables:

$\frac{dy}{\sqrt{y}} = -(3x^2 - 2\sin x) dx$.

Integrate both sides:

$\int \frac{dy}{\sqrt{y}} = \int -(3x^2 - 2\sin x) dx$

$\int y^{-1/2} dy = -\int (3x^2 - 2\sin x) dx$

$2y^{1/2} = -(x^3 + 2\cos x) + C$

$2\sqrt{y} = -x^3 - 2\cos x + C$.

Using the initial condition $y(0) = 1$, we have

$2\sqrt{1} = -0^3 - 2\cos(0) + C$

$2 = -0 - 2(1) + C$

$2 = -2 + C$

$C = 4$.

So, $2\sqrt{y} = -x^3 - 2\cos x + 4$.

$\sqrt{y} = -\frac{x^3}{2} - \cos x + 2$.

Then, we square both sides:

$y = (-\frac{x^3}{2} - \cos x + 2)^2$.

3. Final Answer

$y = \left(-\frac{x^3}{2} - \cos x + 2\right)^2$