91Ó°ÊÓ

In differential equations, an initial value problem (IVP) is where we have not only a differential equation but also an initial condition or a starting point. For example, in our exercise, the differential equation is given as \(x' = t e^{-t^2}\) with the initial condition \(x(0) = 1\). This means we are not only looking for a general function \(x(t)\) that satisfies the differential equation but also one that passes through the point \((0, 1)\).
This initial condition is crucial because there can be infinitely many solutions to a differential equation. The initial condition helps us "pin down" exactly which one corresponds to our problem, ensuring uniqueness of the solution for the given differential equation.

Integration techniques are strategies used to solve integrals, which are essential in finding the solutions to differential equations. Since our differential equation \(x' = t e^{-t^2}\) requires us to integrate the function \(t e^{-t^2}\), understanding which technique to apply is key.
In many cases, you first try to integrate directly. If that doesn’t work, you may employ methods like substitution, by-parts, or recognizing standard integral forms. Our integral, \(\int t e^{-t^2} \, dt\), doesn't fit a basic formula, so other integration techniques like substitution are needed to facilitate the process, allowing us to complete the integration.

The substitution method is a valuable technique in integration, often used when direct integration is complex or challenging. For the integral \(\int t e^{-t^2} \, dt\), we choose the substitution \(u = -t^2\) because the derivative \(du = -2t \, dt\) neatly simplifies the integral.
This transforms our original integral into a simpler form, \(-\frac{1}{2} \int e^{u} \, du\), which is much easier to integrate. By performing this substitution, we transform our challenging problem into a more manageable one. Remember that after integrating with respect to \(u\), it's important to revert back, substituting \(u = -t^2\) into the result to express it in terms of the original variable \(t\).

The general solution of a differential equation is the form that includes all possible solutions, typically containing constants that can be adjusted to fit specific initial conditions. In our case, after integration and substitution, we find the solution \(x(t) = -\frac{1}{2} e^{-t^2} + C\).
The constant \(C\) represents the family of solutions before the initial condition is applied. For an initial value problem like ours, we use the initial condition \(x(0) = 1\) to determine the particular value of \(C\). By solving \(-\frac{1}{2} + C = 1\), we find \(C = \frac{3}{2}\).
Substituting \(C\) into the general solution gives us the specific solution: \(x(t) = -\frac{1}{2} e^{-t^2} + \frac{3}{2}\), which satisfies both the differential equation and the initial condition, completing our process.