91Ó°ÊÓ

According to the Picard-Lindelöf theorem, if \(f(x,y)\) and its partial derivative with respect to y (i.e., \(\frac{\partial f}{\partial y}\)) are continuous on a rectangle R, then the IVP has a unique solution on some interval \([-\alpha, \alpha]\). We have \(f(x,y) = 1+y^2\). Compute the partial derivative with respect to y: $$ \frac{\partial f}{\partial y} = \frac{\partial (1+y^2)}{\partial y} = 2y $$ Since \(2y\) is continuous on R, we can apply Picard-Lindelöf theorem. Now, we find \(\alpha\) by satisfying the Lipschitz continuity: $$ L = \sup |\frac{\partial f}{\partial y}| = \sup |2y| $$ As \(R=\{(x,y)| |x|<10, |\ y-1 \mid<b\}\), we have y in the range \(-b+1\) to \(b+1\). It means \(|2y|\) is maximized when \(|y|=b+1\). Thus we get: $$ L = 2(b+1) $$ The Lipschitz condition states that \(h = min(\frac{b}{L}, 10)\). Using this, we find the interval by satisfying the following condition: $$ -\alpha \le x \le \alpha \text{, with } \alpha = min(\frac{b}{L}, 10) $$

To maximize the interval length \(2\alpha\), we need to maximize \(\frac{b}{L}\). Since \(L = 2(b+1)\), our goal is to maximize $$ \frac{b}{2(b+1)} $$ Taking the derivative of \(\frac{b}{2(b+1)}\) with respect to b, we get: $$ \frac{\partial}{\partial b} (\frac{b}{2(b+1)}) = \frac{2(b+1) - 2b}{[2(b+1)]^2} = \frac{2}{[2(b+1)]^2} $$ As \(\frac{2}{[2(b+1)]^2}\) is always positive, the function \(\frac{b}{2(b+1)}\) increases with increasing b. However, our domain for \(|\ y-1 \mid<b\) restricts b. So, we need to choose the largest possible b for the given domain. As the maximum of x is 10, we can set: $$ b = 10 $$

Now we solve the IVP: $$ y'(x) = f(x, y(x)), \quad y(0) = 1, $$ with \(f(x,y) = 1+y^2\). This is a separable differential equation. Rewrite the equation as $$ \frac{dy}{dx} = 1+y^2. $$ Separate the variables: $$ \frac{dy}{1+y^2} = dx $$ Integrate both sides: $$ \int \frac{dy}{1+y^2} = \int dx $$ $$ \arctan y = x + C $$ Using the initial condition \(y(0) = 1\), we find $$ \arctan 1 = C \Rightarrow C = \frac{\pi}{4} $$ So the solution to the IVP is $$ y(x)=\tan(x+\frac{\pi}{4}) $$ Now, check for the interval of existence. \(y(x) = \tan(x+\frac{\pi}{4})\) is defined for all x such that \(x+\frac{\pi}{4} \neq \frac{\pi}{2} + k\pi\), with \(k\) being an integer. The closest x value that makes the function undefined is \(x = -\frac{3\pi}{4}\). So, the solution exists for the interval: $$ -\frac{3\pi}{4} < x < \frac{3\pi}{4} $$