solve---Find the solution to $\begin{cases} x'(t) = 1 + (x(t))^2 \\ x(0) = 0 \end{cases}$, where $x$ is a real-valued function. In addition, determine its maximal interval of existence.

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We need to solve the differential equation x prime of t equals 1 plus x squared, with initial condition x of 0 equals 0. This is a separable differential equation that we can solve using the separation of variables method. To solve this separable differential equation, we first separate the variables by dividing both sides by 1 plus x squared and multiplying by dt. Then we integrate both sides. The integral of 1 over 1 plus x squared is arctan x, so we get arctan x equals t plus C. Taking the tangent of both sides gives us the general solution: x equals tan of t plus C. Now we apply the initial condition x of 0 equals 0. Substituting into our general solution, we get tan of C equals 0, which means C equals 0. Therefore, our particular solution is x of t equals tan t. We can see this graphically where the curve passes through the origin. The tangent function has vertical asymptotes at odd multiples of pi over 2. Since our initial condition is at t equals 0, the maximal interval of existence is the largest interval containing 0 where the solution remains finite. This gives us the open interval from negative pi over 2 to pi over 2. To solve this separable differential equation, we first separate the variables by dividing both sides by 1 plus x squared and multiplying by dt. Then we integrate both sides. The integral of 1 over 1 plus x squared is arctan x, so we get arctan x equals t plus C. Taking the tangent of both sides gives us the general solution: x equals tan of t plus C. Now we apply the initial condition x of 0 equals 0. Substituting into our general solution, we get tan of C equals 0, which means C equals 0. Therefore, our particular solution is x of t equals tan t. We can see this graphically where the curve passes through the origin. The tangent function has vertical asymptotes at odd multiples of pi over 2. Since our initial condition is at t equals 0, the maximal interval of existence is the largest interval containing 0 where the solution remains finite. This gives us the open interval from negative pi over 2 to pi over 2. In summary, we solved the differential equation x prime equals 1 plus x squared with initial condition x of 0 equals 0. The solution is x of t equals tangent t, with maximal interval of existence from negative pi over 2 to pi over 2. We can verify this solution satisfies both the differential equation and initial condition. This type of equation models exponential growth with finite-time blowup behavior.