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Differential equations are equations involving derivatives of a function. A first-order linear differential equation is a specific type of these equations. It involves the first derivative of a function and has the standard form \( \frac{dy}{dx} + p(x)y = q(x) \). - Here, \( y \) is the function we’re trying to find.- \( p(x) \) and \( q(x) \) are functions of the independent variable. In the exercise, our equation is \( \frac{dB}{dt} + 0.1B = 10 \), making it a first-order linear differential equation since it includes the first derivative \( \frac{dB}{dt} \) and a constant coefficient \( 0.1 \).The goal with such equations is to find a general solution that satisfies it. This is often done by combining the solutions of the corresponding homogeneous equation and a particular solution.

A particular solution addresses only one part of the differential equation problem at hand. It's a specific solution derived from the non-homogeneous counterpart of the original differential equation.- For our equation, the non-homogeneous differential equation is \( \frac{dB}{dt} + 0.1B = 10 \).- To find a particular solution, you usually guess a form based on \( q(x) \) or set coefficients to solve In this exercise, by setting \( B_p = k \), we assume that our particular solution is a constant. When substituting, we get \( 0.1k = 10 \), giving us \( k = 100 \).Thus, our particular solution is \( B_p(t) = 100 \).Particular solutions provide a specific example that satisfies the differential equation.

A homogeneous equation is derived from removing the non-homogeneous part (i.e., \( q(x) \)) of our differential equation. This involves considering\( \frac{dB}{dt} + 0.1B = 0 \).Solutions to this are critical, as they form part of the general solution when combined with the particular solution. Using various methods like separation of variables, the homogeneous solution is often exponential.For our equation, solving gives us \( B_h(t) = Ce^{-0.1t} \), where \( C \) is an arbitrary constant.The form involves exponential functions that decay or grow depending on the signs and values of the coefficients. This aspect allows you to adjust the solution to fit various initial conditions.

Initial conditions allow us to find the specific solution that not only satisfies the differential equation but also meets given criteria at specific points.- In our example, the initial condition is \( B(2) = 3 \).By substituting in the general equation, \( B(t) = Ce^{-0.1t} + 100 \),we solve for \( C \):\( 3 = Ce^{-0.2} + 100 \), leading to \( Ce^{-0.2} = -97 \).Solving this, we find \( C = -97e^{0.2} \).This shows how initial conditions personalize a solution, ensuring that it meets particular criteria at a given point. This step is crucial when applying mathematics to real-world contexts where specific measurements or data points are known.