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A first-order linear differential equation is one of the simplest types of differential equations, yet it's incredibly important.- The standard form of a first-order linear differential equation is \( \frac{dy}{dt} + a(t)y = b(t) \).- These equations involve derivatives of the unknown function, in this case, \(P(t)\), with respect to a single variable, usually time \(t\).The equation given: \[ \frac{dP}{dt} + P = Pt e^{t} \]falls into this category, where \( a(t) = 1 \) and \( b(t) = t e^t \). Recognizing this structure is crucial, as it helps determine the method of solution.Such equations can model numerous real-world processes like cooling, population growth, or radioactive decay. They are critical to understanding how variables change over time or distance under constant rates.

The integrating factor is a magical tool used to solve first-order linear differential equations. It transforms the equation into a more manageable form.- The integrating factor is typically denoted as \( \mu(t) \) and is defined as: \[ \mu(t) = e^{\int a(t) \ dt} \].For our problem, where \( a(t) = 1 \), the integrating factor becomes:\[ \mu(t) = e^{\int 1 \ dt} = e^{t} \].Multiplying the entire differential equation by this integrating factor leads to a perfect derivative on one side:\[ e^{t} \frac{dP}{dt} + e^{t}P = te^{2t} \]This simplifies the equation into a form that can be easily integrated, allowing us to move towards finding the solution.

Integration by parts is a method derived from the product rule for differentiation. It's especially useful when dealing with integrals involving products of functions.The formula for integration by parts is:\[ \int u \, dv = uv - \int v \, du \]In our problem, to integrate: \( \int t e^{2t} \, dt \), we choose:- \( u = t \) (thus \( du = dt \))- \( dv = e^{2t} dt \) (thus \( v = \frac{1}{2} e^{2t} \))Substituting into the formula gives us:\[ \int t e^{2t} \, dt = \frac{t}{2} e^{2t} - \int \frac{1}{2} e^{2t} \, dt \]This helps us break down the complex integral into manageable parts, eventually leading to a solution for the differential equation.

Finding the solution to a differential equation often requires piecing everything together methodically.- After integrating, we obtain an expression for \( e^{t}P \), which in this case is: \[ \frac{t}{2}e^{2t} - \frac{1}{4}e^{2t} + C \]- To isolate \( P(t) \), divide through by \( e^{t} \) yielding: \[ P(t) = \frac{1}{2} te^{t} - \frac{1}{4} e^{t} + Ce^{-t} \]The final step is applying the initial condition to pinpoint the constant \( C \). Here, using \( P(0) = 1 \), we solve:- \( 1 = -\frac{1}{4} + C \),resulting in \( C = \frac{5}{4} \).Plugging back \( C \), the complete solution is:\[ P(t) = \frac{1}{2} te^{t} - \frac{1}{4} e^{t} + \frac{5}{4} e^{-t} \]This solution shows how the function \( P(t) \) behaves with time, given the specific initial condition provided.