Solving 3y'(t) = -15y(t) with y'(0) = 2: A Simple Guide

Today we'll explore a first-order linear differential equation and understand its components.Let's examine our differential equation: three y prime of t equals negative fifteen y of t.Let's break down each component of this equation to better understand what it means.This equation has several important characteristics that make it a first-order linear differential equation.We're also given an initial condition: y prime at t equals zero equals two. This will help us find a unique solution later.The derivative y prime of t represents the rate of change of our function y at any time t.In this equation, the rate of change is proportional to the current value of the function, but with a negative coefficient.Now that we understand the components of our differential equation, we can move on to rearranging it into standard form.Starting with our original equation, three y prime of t equals negative fifteen y of tTo get the standard form, we divide both sides by threeThis simplifies to y prime of t equals negative five y of tThis standard form tells us that the rate of change of y is negative five times the current value of yLet's visualize this with a graph. The negative coefficient indicates that this is an exponential decay equationThe solution curve shows how the value decreases over time, with the rate of decrease being proportional to the current valueAt any point on the curve, the slope of the tangent line is negative five times the y-value at that pointNotice how the slope becomes less steep as the value of y decreases, showing that the rate of decay slows down over timeThe general solution to our differential equation takes the form of an exponential function.We can write this as y of t equals C times e to the negative five t.Let's verify this is indeed a solution by following these steps.First, we start with our proposed solution.When we differentiate this, we get negative five C e to the negative five t.This equals negative five times our original function, exactly matching our differential equation.The constant C determines which specific solution we get. Let's look at some examples.When C equals two, the curve starts higher but follows the same decay pattern.With a negative value of C, the curve starts below zero but still decays exponentially.The key feature of this solution is that its derivative equals negative five times itself, which is exactly what our differential equation requires.Now that we have our general solution, we can use the initial condition to find the specific value of C.Now that we have our general solution, we need to find the specific value of C using our initial condition.We know that the derivative at time zero equals two.Let's substitute t equals zero into our derivative equation.When we substitute zero, e to the negative five times zero equals one.This simplifies to negative five C equals two.Solving for C, we divide both sides by negative five, giving us C equals negative two fifths.Let's verify our solution by plugging C equals negative two fifths back into the derivative equation at t equals zero.This value of C gives us our unique solution that satisfies both the differential equation and the initial condition.Now that we have found C equals negative two fifths, we can write our final solution.Substituting this value into our general solution gives us y of t equals negative two fifths e to the negative five t.This solution represents an exponential decay function. As time increases, the function approaches zero, but never quite reaches it.Let's verify that our solution satisfies both the original differential equation and the initial condition.When we take the derivative of our solution, we get two e to the negative five t, which at t equals zero gives us our initial condition of two.Our solution has several key characteristics that make it unique and verify it's the correct answer to our differential equation.