91Ó°ÊÓ

In the realm of first-order linear differential equations, the integrating factor is a magic multiplier that simplifies solving the equation. For the differential equation \( \frac{\mathrm{d}x}{\mathrm{d}t} - tx = -2t \), the integrating factor helps us to reframe and solve it efficiently. But what exactly is it?

Simply put, an integrating factor is a function, here noted as \( \mu(t) \), which is commonly derived from the exponentiation of the integral of the coefficient of \( x \) in the standard form of the equation. For our equation, the coefficient of \( x \) is \( -t \). So we calculate the integral as \( \int -t \, \mathrm{d}t = -t^2/2 \).

• The integrating factor \( \mu(t) \) is then given by \( e^{-t^2/2} \).
• This transforming step ensures we can express the left-hand side of our rearranged equation as a derivative.

Multiplying the whole equation by this integrating factor converts it into a form that is easier to integrate, leading us smoothly toward the solution.

The general solution of a differential equation provides us a family of solutions that encompass all particular instances of the answer. For our equation, after establishing our integrating factor \( e^{-t^2/2} \), we rewrite the equation as a derivative, which simplifies the solving process.

Consider the left side of the modified equation: it becomes \( \frac{\mathrm{d}}{\mathrm{d}t}(e^{-t^2/2}x) \). This nifty trick makes it straightforward to integrate both sides. After performing the integration, we deduce:

• The left side integrates to \( e^{-t^2/2}x \).
• The right side, after some substitution and manipulation, simplifies down to \( 2e^{-t^2/2} + C \), where \( C \) is an arbitrary constant.

To solve for \( x(t) \), we multiply through by \( e^{t^2/2} \), leading to the general solution:
\( x(t) = 2 + Ce^{t^2/2} \). This result includes all specific solutions depending on the constant \( C \).

Particular solutions provide specific answers for differential equations based on given initial conditions. Unlike the general solution, which spans many possible solutions, a particular solution satisfies a specific point. For this reason, we use initial conditions like \( x(0) = 5 \) to pinpoint this unique solution.

To find the particular solution, substitute the initial condition into the general solution \( x(t) = 2 + Ce^{t^2/2} \). Here's how:

• Set \( t = 0 \) in the general solution, yielding \( x(0) = 2 + Ce^0 = 2 + C \).
• Equate this with the provided initial condition \( x(0) = 5 \).

Solving \( 5 = 2 + C \) gives \( C = 3 \). Substituting back, the particular solution becomes
\( x(t) = 2 + 3e^{t^2/2} \). This delivers a precise solution that passes through the initial condition point, offering clarity and specificity.