Finding Particular Solution to Differential Equation

General and Particular Solutions Here we will learn to find the general solution of a differential equation, and use that general solution to find a particular solution. We will also apply this to acceleration problems, in which we use the acceleration and initial conditions of an object to find the position function. Example 1: Finding a Particular Solution

       Find the particular solution of the differential equation which satisfies the given inital condition:

       First, we need to find the general solution. To do this, we need to integrate both sides to find y:

       This gives us our general solution. To find the particular solution, we need to apply the initial conditions given to us (y = 4, x = 0) and solve for C:

       After we solve for C, we have the particular solution.

Example 2: Finding a Particular Solution

       Find the particular solution of the differential equation which satisfies the given inital condition:

       First, we need to integrate both sides, which gives us the general solution:

       Now, we apply the initial conditions (x = 1, y = 4) and solve for C, which we use to create our particular solution:

Example 3: Finding a Particular Solution

       Find the particular solution of the differential equation which satisfies the given inital condition:

       First, we find the general solution by integrating both sides:

       Now that we have the general solution, we can apply the initial conditions and find the particular solution:

Velocity and Acceleration Here we will apply particular solutions to find velocity and position functions from an object's acceleration.

Example 4: Finding a Position Function

       Find the position function of a moving particle with the given acceleration, initial position, and initial velocity:

       We have an acceleration function, an initial velocity of 10, and an initial position of 5, and are looking for the position function. We know that the integral of acceleration is velocity, so let's start with that:

       We now have the general solution of the velocity function. To get the particular solution, we need the initial velocity. Since this is the initial velocity, it is the velocity at time t = 0; therefore, our initial condition is v = 10, t = 0:

       Now that we have the particular solution of velocity, we can integrate it to find position:

       Now we can apply our initial conditions to this general solution to get the particular solution, which is the position function that we want. Just like before, x ₀ is theinitial
position, which means that the time t = 0, and x = 5:

       This is the position function of the particle.

Example 5: Finding a Position Function

       Find the position function of a moving particle with the given acceleration, initial position, and initial velocity:

       We have an equation for acceleration, an initial velocity of 7, and an initial position of 0. The first step is finding the particular solution of the particle's velocity:

       Now, we can use the velocity function to find the position function. Remember, we will need to find the particular solution of the position function, not just the general solution:

Example 6: Applying a Differential Equation

       Here, we will use a real-life example to apply what we just learned.

       A ball is thrown straight downward with an initial speed of 20 ft/s from the top of a building which is 300 feet tall. Ignoring air friction, dow long does it take the ball to reach the ground, and at what speed does it hit?

       To solve this, we need to put it into terms we can understand. The units are given in feet and feet/second; the acceleration due to gravity in these units is -32 ft/s².

       We know that the ball was thrown with an initial speed (t = 0) of 20 ft/s downward; since it is going down, the velocity will be a negative value (v ₀ = -10).

       Finally, the building is 300 feet tall, and the ball is thrown from the top of it. Since the ball starts out at a place upward from ground level, the initial position will be positive 300 (x ₀ = 300). Let's put all this into an equation similar to the previous examples:

       Now we're getting somewhere! The question asks about the ball when it hits the ground. To be able to figure out information for when it hits the ground, we need to know what time it hits. The equation which relates position to time is the position function, which we already know how to get from the previous examples:

       Now that we have the position function, we can start solving for the time it takes the ball to hit the ground, and the velocity at which it hits. Each of these equations needs to know a time; for example, if we plug in 2 for t in the velocity function, it will give us the velocity at t = 2, or 2 seconds after the ball is thrown. We need to know what time the ball hits the ground; to do this, we need to set the position function equal to 0 and solve for t. The ball started 300 feet from the ground, and we used 300 as our initial position. If we set our position equal to 0, that will tell us when the ball hits the ground:

       We get two values for t: -5 and 3.75. We can throw out the -5, since we can't have a negative value for time. Therefore, the time it takes the ball to reach the ground is 3.75 seconds. To find the velocity when the ball hits the ground, we simply plug in 3.75 for t in our velocity equation and solve:

       The velocity of the ball as it hits the ground is -140 ft/s

Example 7: Applying a Differential Equation

       A car's brakes are applied when it is traveling at 60 km/h, providing a constant deceleration of 12 m/s². How far does the car travel before coming to a stop, and how long does it take?

       Okay, let's break this down. We know the acceleration is -12 m/s². The initial velocity is 60 km/h; this will need to be converted into m/s (we can't have a problem with unlike units):

       The car's initial velocity is 16.7 m/s. We can also call the initial position x = 0, since this is when the car starts decelerating. Putting all this together:

       We know that we will need the position function at some point, since we need to figure out how far the car travels before coming to a stop, so let's go ahead and get that out of the way:

       Now, we need to figure out at what time the car comes to a stop. We don't know what the position of the car will be at this point, but we do know that the velocity will be 0. To find out what time the velocity is 0, we need to set velocity equal to 0 and solve:

       The car comes to a stop 1.4 seconds after applying the brakes. How far does it travel before it stops? We need to plug in t = 1.4 to the position function to find out:

       The car travels 11.6 meters before stopping

Finding Particular Solution to Differential Equation

Source: http://www.copingwithcalculus.com/general-and-particular-solutions.html

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