Solutions (level 1) One of the most important goals in an introductory level Of course, to solve, or find solutions of differential equations. In this video we will explore the concept of a solution of an ordinary differential equation equation. Let us first begin by replacing the following definition: Any function defined on an interval I and owns at least N derivatives that are continuous on I, which then replaced in a nth-order ordinary differential equations (also known as an ODE) reduce the equation around an identity, It is said that a solution of the equation to the interval I. In other words, a solution of a de-order ordinary differential equation is a function that has at least N derivatives and for which the equation equals 0. For all x in I. If this is the case we say that the function satisfies the equator comparison on I. In these videos so we will also assume that a solution is a real-valued function, as opposed to an imaginary or complex function. It can be confusing to use phi to represent the solution of an ODE at a given interval. So we will also use y (x) to indicate a solution of an ODE at a given interval.
It's important to get knowledge right off the bat, that a solution to a differential equation is a function, as opposed to the solution to an algebraic equation which is (usually) a number, or a set of numbers. It makes differential equations much more interesting, and often more challenging to understand, than algebraic equations. It is very important to associate a solution of an ordinary differential equation with an interval. The interval I is known as the interval of definition, it is also referred to as the interval of existence, the interval of validity, or the domain of the solution. This interval can be an open interval, a closed interval, and infinite interval, or all real numbers. For example, say we are asked to verify if the following function y = (1/16) * x ^ 4 is a solution to the given differential equation y '= x * y ^ (1/2), on the interval negative infinity to positive infinity in other words all real numbers.
One way to verify that the given function is a solution is to look at the replacement, or both sides of the equation are the same for each value of x in the interval. Let's first find the derivative of the function take the derivative of 1/16 times x raised to the power of 4 we get, 1/4 times x elevated to the power of 3, You must be comfortable take derivatives if you need a refresher course you can always go back look at the derivatives videos under Calculus I.
Well after found the derivative can continue to replace the function y and its derivative. always use brackets when performing a replacement this will help reduce any errors associated with negative signs. Performing the replacement and the reduction of the expression we get the following, realize that we reduce the equation in an identity, each side of the equation is the same for each real number x in the interval. This means that the function is indeed a solution to the differential equation on the given interval in this case all real numbers. At times the interval may not all be realistic numbers as in the following example. Show that the following function y = x ^ (- 3/2) is a solution to the differential equation 4x ^ 2 * y '' + 12 * x * y '+ 3y = 0, for x> 0, Good look at the ODE we see that we go to the first and second derivative of need the function.
The first derivative equal to the next expression and second derivative is equal to the next, it's just a matter of the application of the power rule of calculus I or differential calculus. Next we continue and replace the function and its derivative in the ODE as follows, to make sure we use brackets when performing the replacement do what we get next, next we goes on and the expressions do simplify that we get the following, notice that We can simplify the left side of the equation by collecting similar terms, do it We get the following, 0 = 0 notice that the left side is equal to the right hand side of the equation so the function is a solution of the differential equation, we has not yet done what we need to show an interval for this solution, notice that the interval has already been given to us in this case the interval is from 0 exclusive to positive infinity or all the values of x that are greater than is 0.
Why is this function a solution to the ODE in just this interval? Remember that we can rewrite the solution for this ODE is as follows: 1 cut into squares over the square root of x. In this form it is clear that we need to avoid x = 0 because division by zero is undefined. We also have a radical function specifically a square root function. Remember that we are only interested in real values, so if our differential equation only contains real numbers then what we do not want solutions that produce imaginary or complex numbers. So, in an effort to avoid imaginary or complex numbers we will also need to avoid negative values of x. As a result, our domain is limited to only positive values that is greater than 0, hence the interval of definition. This is why it is important to define an interval when finding a solution a differential equation. Even if a function can symbolically satisfy a differential equation, due to certain limitations brought about by the solution We can not use all the values of the independent variable and therefore must make a constraint on the independent variable.
It will be so the case with many solutions for differential equations. Note that each differential equation has the constant solution y = 0 in the interval negative infinity to positive infinity or all real numbers. If we replace this function and its derivative in the ODE would technically comply with the ODE. A solution of a differential equation that is identical to zero at an interval I is said be a trivial solution. For the most part we will want to find non-trivial solutions from ODE. With that in mind, we will shift gears and talk about solution curves. The graph of a solution of an ODE is known as a solution curve.
Since the solution is a differentiable function, it is continuous on its interval I of definition. With that said, there may be a difference between the graph of the function and the graph of the solution. In other words, the domain of the function is not always the same as the interval I of definition (or domain) of the solution. Let us illustrate this with the following example. Make sure that the function y = 1 / x is a solution to the differential equation xy '+ y = 0. Good notice that this ODE is a first-order linear ODE, so we need to replace the function and the first derivative in the ODE. Let's first find the derivative of the function, the application of the power rule we obtain the following, next we proceed and replace the function and its derivative in the equation as follows, then we go ahead and simplify the expressions, then we get to do similar terms we get the following expression 0 = 0, so we verified that this function is indeed a solution for the ODE, now we need to assign an appropriate interval of definition, to find out the interval of definition Let's take a look at the graph of y = 1 / x.
Note that the domain of this function is the collection of all real numbers x except 0. The rational function is discontinued at x = 0, it has essentially a vertical asymptote at x = 0. Due to this discontinuity the function not differentiable at x = 0, remember that a solution for a differential equation have been differentiable in the given interval as well as the satisfaction of the differential equation. If we look at this function as a solution to the differential equation we should show an interval of definition, this interval can be anywhere except at x = 0. It can be the interval [-2, -1], or the interval [3, 4], or the interval of negative infinity after 0 exclusive, or the interval of 0 exclusive to positive infinity. In all these intervals the function is differentiable and produces real numbers, as opposed to complex or imaginary numbers. For the most part we want the interval of definition to be as large as possible so that We want me to be either negative infinity to 0 exclusive or 0 exclusive to positive infinity. In this example it does not really matter where we choose interval as long as the solution is differentiable in the given interval and yield real numbers.
In a much later video we will encounter specific issues we need to choose from one interval over the other for now either interval are acceptable answers. Well in our next video we will practice verifies solutions for differential equations and the determination of appropriate intervals of definition..