91Ó°ÊÓ

In the realm of first-order linear differential equations, finding the integrating factor is a crucial step. The integrating factor transforms the equation into a form that is easier to solve. For a differential equation in the form \( y' + P(x)y = Q(x) \), the integrating factor \( \mu(x) \) is expressed as: \[ \mu(x) = e^{\int P(x) \, dx} \]. This effectively allows us to rewrite the equation in a manner that highlights the derivative of a product on the left-hand side. In our exercise, with \( P(x) = 1 \), this simplifies to \( \mu(x) = e^x \). As a result, multiplying through by the integrating factor, we get:

• \( e^x y' + e^x y = e^{x^2} \)

This transformation is pivotal, making the left side resemble the derivative \( \frac{d}{dx}(e^x y) \). It turns our differential equation into an integrable form. By identifying and utilizing the integrating factor correctly, we pave the way for a straightforward integration process to unfold.

Initial conditions in differential equations serve as essential tools for arriving at a specific solution from the general solution family. They give us particular values to satisfy at a specific point \( x = a \). Here, the initial condition \( y(0) = 10 \) helps us find the constant of integration, \( C \), necessary to craft our specific solution.After reaching the stage: \[ e^x y = \int e^{x^2} \, dx + C \], substituting the initial condition, we have:

• \( e^0 y(0) = \int_0^0 e^{t^2} \, dt + C = 10 \).

Evaluating this expression, the term \( \int_0^0 e^{t^2} \, dt \) yields 0 (since the bounds of the integral are the same), revealing that \( C = 10 \). With the initial condition applied, our particular solution refines, ensuring that the function is correctly constrained to meet the conditions given at \( x = 0 \). Initial conditions infuse our solution with precision, rooting it in reality by aligning it with known values.

A definite integral provides us with a specific numerical value accumulated over an interval. In this exercise, the transformation process of the differential equation leads us to express the solution using a definite integral. We integrate from \( 0 \) to \( x \), capturing the specific area under the curve, embodied as: \[ y(x) = e^{-x} \left( \int_0^x e^{t^2} \, dt + 10 \right) \] The definite integral \( \int_0^x e^{t^2} \, dt \) captures the accumulation from the start point (at \( 0 \)) to a variable endpoint (\( x \)), accounting for changes due to the exponent inside the integrand. Although, in this case, the integral \( \int e^{x^2} \) lacks a closed form, representing it as a definite integral is a valid and powerful method in mathematics. This allows us to sidestep the intractability, instead expressing the response exactly in terms of an integral, which accommodates a wide range of practical computations or numerical methods for evaluation.