Solving Differential Equations Using Integrating Factor
Dodie Clymore
The first special case of first order differential equations that we will look at is the linear first order differential equation. In this case, unlike most of the first order cases that we will look at, we can actually derive a formula for the general solution. The general solution is derived below. However, we would suggest that you do not memorize the formula itself. Instead of memorizing the formula you should memorize and understand the process that I'm going to use to derive the formula. Most problems are actually easier to work by using the process instead of using the formula.
So, let's see how to solve a linear first order differential equation. Remember as we go through this process that the goal is to arrive at a solution that is in the form \(y = y\left( t \right)\). It's sometimes easy to lose sight of the goal as we go through this process for the first time.
Where both \(p(t)\) and \(g(t)\) are continuous functions. Recall that a quick and dirty definition of a continuous function is that a function will be continuous provided you can draw the graph from left to right without ever picking up your pencil/pen. In other words, a function is continuous if there are no holes or breaks in it.
Now, from a notational standpoint we know that the constant of integration, \(c\), is an unknown constant and so to make our life easier we will absorb the minus sign in front of it into the constant and use a plus instead. This will NOT affect the final answer for the solution. So with this change we have.
Again, changing the sign on the constant will not affect our answer. If you choose to keep the minus sign you will get the same value of \(c\) as we do except it will have the opposite sign. Upon plugging in \(c\) we will get exactly the same answer.
There is a lot of playing fast and loose with constants of integration in this section, so you will need to get used to it. When we do this we will always to try to make it very clear what is going on and try to justify why we did what we did.
Now, the reality is that \(\eqrefeq:eq9\) is not as useful as it may seem. It is often easier to just run through the process that got us to \(\eqrefeq:eq9\) rather than using the formula. We will not use this formula in any of our examples. We will need to use \(\eqrefeq:eq10\) regularly, as that formula is easier to use than the process to derive it.
Note that officially there should be a constant of integration in the exponent from the integration. However, we can drop that for exactly the same reason that we dropped the \(k\) from \(\eqrefeq:eq8\).
From this point on we will only put one constant of integration down when we integrate both sides knowing that if we had written down one for each integral, as we should, the two would just end up getting absorbed into each other.
The final step in the solution process is then to divide both sides by \(\bfe^0.196t\) or to multiply both sides by \(\bfe^ - 0.196t\). Either will work, but we usually prefer the multiplication route. Doing this gives the general solution to the differential equation.
From the solution to this example we can now see why the constant of integration is so important in this process. Without it, in this case, we would get a single, constant solution, \(v(t)=50\). With the constant of integration we get infinitely many solutions, one for each value of \(c\).
Back in the direction field section where we first derived the differential equation used in the last example we used the direction field to help us sketch some solutions. Let's see if we got them correct. To sketch some solutions all we need to do is to pick different values of \(c\) to get a solution. Several of these are shown in the graph below.
Now, recall from the Definitions section that the Initial Condition(s) will allow us to zero in on a particular solution. Solutions to first order differential equations (not just linear as we will see) will have a single unknown constant in them and so we will need exactly one initial condition to find the value of that constant and hence find the solution that we were after. The initial condition for first order differential equations will be of the form
To find the solution to an IVP we must first find the general solution to the differential equation and then use the initial condition to identify the exact solution that we are after. So, since this is the same differential equation as we looked at in Example 1, we already have its general solution.
Now, to find the solution we are after we need to identify the value of \(c\) that will give us the solution we are after. To do this we simply plug in the initial condition which will give us an equation we can solve for \(c\). So, let's do this
Can you do the integral? If not rewrite tangent back into sines and cosines and then use a simple substitution. Note that we could drop the absolute value bars on the secant because of the limits on \(x\). In fact, this is the reason for the limits on \(x\). Note as well that there are two forms of the answer to this integral. They are equivalent as shown below. Which you use is really a matter of preference.
This is an important fact that you should always remember for these problems. We will want to simplify the integrating factor as much as possible in all cases and this fact will help with that simplification.
Now back to the example. Multiply the integrating factor through the differential equation and verify the left side is a product rule. Note as well that we multiply the integrating factor through the rewritten differential equation and NOT the original differential equation. Make sure that you do this. If you multiply the integrating factor through the original differential equation you will get the wrong solution!
In Maths, an integrating factor is a function used to solve differential equations. It is a function in which an ordinary differential equation can be multiplied to make the function integrable. It is usually applied to solve ordinary differential equations. Also, we can use this factor within multivariable calculus. When multiplied by an integrating factor, an inaccurate differential is made into an accurate differential (which can be later integrated to give a scalar field). It has a major application in thermodynamics where the temperature becomes the integrating factor that makes entropy an exact differential.
Integrating factor is defined as the function which is selected in order to solve the given differential equation. It is most commonly used in ordinary linear differential equations of the first order.
Integrating factor and Laplace are two different mathematical techniques used in solving engineering problems. Integrating factor is used to solve linear first-order differential equations, while Laplace is used to solve linear differential equations of any order. Additionally, integrating factor involves multiplying the entire equation by a factor, while Laplace involves transforming the equation into a different domain.
The choice between integrating factor and Laplace depends on the type of differential equation you are trying to solve. If the equation is linear and first-order, you should use integrating factor. If the equation is linear and of any order, you should use Laplace. It is important to note that Laplace can also be used for non-linear equations, while integrating factor cannot.
Yes, integrating factor and Laplace can be used together to solve a differential equation. In some cases, using both techniques may be necessary to fully solve the problem. However, it is important to carefully consider the problem and determine which technique is most appropriate before attempting to combine them.
Both integrating factor and Laplace have their own limitations. Integrating factor can only be used for linear first-order equations, while Laplace can only be used for linear equations. Additionally, Laplace may not work for some non-linear equations, and integrating factor may not be efficient for higher-order equations. It is important to understand the strengths and limitations of each technique in order to choose the most appropriate one for a given problem.
The choice between integrating factor and Laplace can greatly affect the solution of an engineering problem. Due to their different approaches, the solutions obtained from integrating factor and Laplace may differ in form and complexity. In some cases, one technique may be more efficient or easier to use than the other. It is important to carefully consider the problem and choose the most appropriate technique in order to obtain an accurate and efficient solution.
As background, recall one of the simplest types of ODEs mentioned in the introductory page. If we have an equation such as $$\diffxt = t^2,$$ we can quickly solve it by integration. This equation is so simple because the left hand side is just a derivative with respect to $t$ and the right hand side is just a function of $t$. We can solve by integrating both sides with respect to $t$ to get that $x(t)=\fract^33 + C$.
For this choice of $\mu$, we can exchange $\mu$ with the equivalent expression $\diff\mut$, and the lefthand side of equation \eqrefmultmu1 is indeed the derivative of $\mu(t)x(t)$.We can rewrite the equation as \begingather* \difft(C_1e^tx(t))=C_1e^tt^2. \endgather*
The new version of our ODE is \begingather* \difft(e^tx(t))=e^tt^2. \labelodederiv1\tag4 \endgather* Finally we have transformed the ODE of \eqrefode1 to the simple form we desired. The left hand side of equation \eqrefodederiv1 is a derivative with respect to $t$ and the right hand side is a function of $t$ alone. We can find the solution by integrating with respect to $t$: \beginalign* \int \difft(e^tx(t)) dt &= \int e^tt^2 dt + C\\ e^t x(t) &= e^t (2 - 2 t + t^2) +C. \endalign*
In this case, the integral $\int e^tt^2 dt$ was simple enough that we could calculate the result analytically by integrating by parts two times to obtain $e^t (2 - 2 t + t^2)$. Even if we ended up with an integral that we couldn't compute, we would still consider the ODE to be solved, leaving the solution in terms of an integral.
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