Linear differential equations of the first order
Linear differential equations of the first order |
Definition of a first-order linear equation Differential equation of the form ![]() where a ( x ) and b ( x ) are continuous functions of x, are called linear nonhomogeneous differential equations of the first order. We will consider two methods for solving these equations:
Using an integrating factor If a linear differential equation is written in standard form: ![]() then the integrating factor is determined by the formula: ![]() Multiplying the left-hand side of the equation by the integrating factor u ( x ) transforms it into the derivative of the product y ( x ) u ( x ). The general solution of the differential equation is expressed as: ![]() where C is an arbitrary constant. Method of variation of constant This method is similar to the previous approach. First, it is necessary to find the general solution of the homogeneous equation : ![]() The general solution of the homogeneous equation contains a constant of integration C. Next, we replace the constant C with some (as yet unknown) function C ( x ). By substituting this solution into the inhomogeneous differential equation, we can determine the function C ( x ). The described algorithm is called the method of variation of the constant. Of course, both methods lead to the same result. Cauchy problem If, in addition to the differential equation, an initial condition of the form y ( x 0 ) = y 0 is also given, then such a problem is called a Cauchy problem. The solution to the Cauchy problem does not contain an arbitrary constant C. Its specific numerical value is determined by substituting the general solution of the equation into the given initial condition y ( x 0 ) = y 0. |
Example 1 |
Solve the equation y’ − y − xe x = 0. Solution. Let us write this equation in standard form: ![]() We will solve this equation using an integrating factor: ![]() Then the general solution of the linear differential equation is determined by the expression: ![]() |
Example 2 |
Solve the differential equation Solution. We’ll solve this problem using the method of variation of a constant. First, we’ll find the general solution to the homogeneous equation: ![]() which is solved by separation of variables: ![]() where C is an arbitrary positive number. Now we replace the constant C with some (as yet unknown) function C (x), nd then we will seek a solution to the original inhomogeneous equation in the form: ![]() The derivative is equal to ![]() Substituting this into the differential equation, we obtain: ![]() By integrating, we find the function C ( x ) : ![]() where C 1 is an arbitrary real number. Thus, the general solution of the given equation is written as: ![]() |
Example 3 |
Solve the equation y’ − 2 y = x. Solution. A. First, let’s solve this problem using the integrating factor. Our equation is already written in standard form. Therefore: ![]() Then the integrating factor has the form: ![]() The general solution of the original equation is written as: ![]() Let us calculate the last integral using integration by parts. ![]() We receive ![]() B. Now we will construct a solution using the method of variation of the constant. Let us consider the corresponding homogeneous equation ![]() and we will find its general solution: ![]() where C again denotes an arbitrary real number. Note that when C = 0, we obtain the solution y = 0, which also satisfies the homogeneous equation. Next, assume that C is a function of x and substitute the solution y = C ( x ) e 2 x into the original inhomogeneous equation. The expression for the derivative is: ![]() Hence, ![]() This integral was already found in point A, so we can write: ![]() As a result, the general solution of the inhomogeneous differential equation is expressed by the formula: ![]() As you can see, both methods lead to the same answer :). |
Example 4 |
Solve the differential equation x 2 y’ + xy + 2 = 0 . Solution. We’ll solve this example using the method of variation of a constant. For convenience, we’ll write the equation in standard form: ![]() Let’s divide both sides by x 2. Obviously, the root x = 0 is not a solution to the equation. Let’s consider the homogeneous equation: ![]() After simple transformations, we obtain the answer: y = C / x, where C is an arbitrary real number. The last expression includes the case y = 0, which is also one of the solutions of the homogeneous equation. Now we replace the constant C with the function C ( x ) and substitute the solution y = C ( x )/ x into the original nonhomogeneous differential equation. Since ![]() then we get: ![]() Thus, the general solution of the original equation is: ![]() |
Example 5 |
Solve the Cauchy problem: Solution. First, we calculate the integrating factor, which is written as ![]() Here ![]() Therefore, the integrating factor is determined by the formula: ![]() We can take the function u ( x ) = cos x as the integrating factor. It is easy to verify that the left-hand side of the equation, after multiplying by the integrating factor, becomes the derivative of the product y ( x ) u ( x ) : ![]() Then the general solution of the given equation is written as follows: ![]() Now we define a constant C that satisfies the initial condition y (0) = 1 : ![]() It follows that C = 4/3. Consequently, the solution to the Cauchy problem is expressed by the formula: ![]() |
Example 6 |
Solve the differential equation (Cauchy problem) Solution. Let’s define the integrating factor: ![]() As such a multiplier, we choose the function u ( x ) = x 3. It can be verified that the left-hand side of the equation after multiplication by the integrating factor will be the derivative of the product y ( x ) u ( x ) : ![]() The general solution of the equation is written in the form: ![]() Now we can find the constant C using the initial condition y (1) = 2. Substituting the general solution into the initial condition gives the following result: ![]() So, the solution to the Cauchy problem looks like this: ![]() |
Example 7 |
Find the general solution of the differential equation y = (2 y 4 + 2 x ) y’. Solution. It is clear that this equation is not linear with respect to the function y ( x ). However, we can try to find a solution for the inverse function x ( y ). Let’s write the given equation in terms of differentials and make some transformations: ![]() We have obtained a linear differential equation with respect to the function x ( y ). Let’s solve it using an integrating factor: ![]() The general solution in the form of an inverse function x ( y ) is expressed by the formula: ![]() |
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