Equations with separable variables
Equations with separable variables |
A first-order differential equation y’ = f ( x,y ) is called an equation with separable variables if the function f ( x,y ) can be represented as a product of two functions depending only on x and y : ![]() where p ( x ) and h ( y ) are continuous functions. Considering the derivative y’ as a ratio of differentials ![]() Of course, we need to make sure that h ( y ) ≠ 0 . If there is a number x 0 such that h ( x 0 ) = 0, then this number will also be a solution to the differential equation. Dividing by h ( y ) results in the loss of the indicated solution. Denoting ![]() Now the variables are separated, and we can integrate the differential equation: ![]() where C is the constant of integration. Calculating the integrals, we obtain the expression. ![]() describing the general solution of an equation with separable variables. |
Example 1 |
Solve the differential equation .
Solution. In this case, p ( x ) = 1 and h ( y ) = y ( y +2) . We divide the equation by h ( y ) and move dx to the right side: ![]() Note that when dividing, we could lose the solutions y = 0 and y = −2 in the case where h ( y ) is equal to zero. Indeed, let us verify that y = 0 is a solution to this differential equation. Let ![]() Substituting this into the equation yields: 0 = 0. Therefore, y = 0 will be one of the solutions. Similarly, we can verify that y = -2 is also a solution to the equation. Let’s return to the differential equation and integrate it: ![]() The integral on the left-hand side can be calculated using the method of undetermined coefficients: ![]() Thus, we obtain the following expansion of the rational fraction in the integrand: ![]() Hence, ![]() Let’s rename the constant: 2 C = C 1 . As a result, the final solution to the equation is written as: ![]() The general solution here is expressed implicitly. In this example, we can transform it and obtain an explicit answer as a function y = f ( x, C 1 ), where C 1 is a constant. However, this cannot be done for all differential equations. |
Example 2 |
Solve the differential equation .
Solution. Let us write this equation in the following form: ![]() Divide both sides by ( x 2 + 4) y : ![]() Clearly, x 2 + 4 ≠ 0 for all real x. Let’s check that y = 0 is one of the solutions to the equation. After substituting y = 0 and dy = 0 into the original differential equation, it is clear that the function y = 0 is indeed a solution to the equation. Now we can integrate the resulting equation: ![]() Note that dx 2 = d ( x 2 + 4) . Therefore, ![]() Let us represent the constant C as ln C1, where C 1 > 0. Then ![]() Thus, the given differential equation has the following solutions: ![]() The answer we obtained can be simplified. In fact, we introduce an arbitrary constant C, taking values from −∞ to +∞. Then the solution can be written as: ![]() When C = 0, it becomes equal to y = 0. |
Example 3 |
Find all solutions of the differential equation y’ = −xe y.
Solution. We transform the equation as follows: ![]() Dividing by e y does not lead to a loss of solutions, since e y > 0. After integration, we obtain ![]() This answer can be expressed explicitly: ![]() In the last expression, the constant C is assumed to be > 0 in order to satisfy the domain of the logarithmic function. |
Example 4 |
Find a particular solution of the differential equation subject to the condition y (1) = −1.
Solution. Let’s divide both sides of the equation by x : ![]() We assume that x ≠ 0, since the domain of the original equation is the set x > 0. As a result of integration, we obtain: ![]() The integral on the right-hand side is calculated as follows: ![]() Therefore, the general solution in implicit form is: ![]() where C 1 = 2 C is the constant of integration. Let us now find the value of C 1 that satisfies the initial condition y (1) = −1 : ![]() Thus, a particular solution of a differential equation with a given initial condition (the Cauchy problem) is described by an algebraic equation: ![]() |
Example 5 |
Solve the differential equation y’ cot 2 x + tan 2 y = 0.
Solution. Let us write this equation in the following form: ![]() Divide both sides by tan y cot 2 x : ![]() Let’s check if we’ve lost any solutions as a result of the division. We need to examine the following two roots: ![]() Substituting into the original equation, we see that ![]() Here we get the answer: ![]() which does not satisfy the original differential equation. Now we can integrate the differential equation and find its general solution: ![]() The final answer is written as: ![]() |
Example 6 |
Find a particular solution of the equation that satisfies the initial condition y (0) = 0.
Solution. Let’s rewrite the equation as follows: ![]() Divide both sides by 1 + e x : ![]() Since 1 + e x > 0, we didn’t lose any solutions during the division. We integrate the resulting equation: ![]() Now we find the constant C from the initial condition y (0) = 0. ![]() Therefore, the final answer is: ![]() |
Example 7 |
Solve the equation .
Solution. The product of xy in each part does not allow for separation of variables. Therefore, we will make a substitution: ![]() The relationship for differentials is: ![]() Substituting this into the equation, we get: ![]() Next, multiplying both parts of x, we can write, after the appropriate abbreviations: ![]() Let’s take into account that x = 0 is a solution to the equation (this can be verified by direct substitution). The last expression can be simplified somewhat: ![]() Now the variables x and t are separated: ![]() As a result of integration, we find: ![]() By performing the back substitution t = xy, we obtain the general solution of the differential equation: ![]() The full answer is written as: ![]() |
Example 8 |
Find the general solution of the differential equation .
Solution. Let’s use the following substitution: ![]() As a result, the equation takes the form: ![]() Hence, ![]() Let’s integrate the last equation: ![]() Since u = x + y, the final answer is implicitly written as: ![]() |
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