Homogeneous equations
Homogeneous equations |
Definition of a homogeneous differential equation First-order differential equation ![]() is called homogeneous if the right-hand side satisfies the relation ![]() for all values of t. In other words, the right-hand side must be a homogeneous function of order zero with respect to the variables x and y : ![]() A homogeneous differential equation can also be written as ![]() or through differentials: ![]() where P ( x,y ) and Q ( x,y ) are homogeneous functions of the same order. Definition of a homogeneous function A function P (x, y) is called a homogeneous function of order n if the following relation holds for all t > 0 : ![]() Solution of homogeneous differential equations A homogeneous differential equation can be solved by substituting y = ux, which transforms the homogeneous equation into an equation with separable variables. A differential equation of the form ![]() is transformed into an equation with separable variables by moving the origin of the coordinate system to the intersection point of the straight lines specified in the equation. If the specified lines are parallel, then the differential equation is reduced to an equation with separable variables by changing the variable: ![]() |
Example 1 |
Solve the differential equation Solution. It is easy to see that the polynomials P ( x,y ) and Q ( x,y ), respectively, for dx and dy, are homogeneous functions of the first order. Therefore, this differential equation will also be homogeneous. Let y = ux, where u is some new function depending on x. Then ![]() Substituting this into the differential equation, we obtain ![]() Hence, ![]() Let’s divide both sides of the equation by x : ![]() By dividing by x, we could have lost the solution x = 0. Direct substitution shows that x = 0 is indeed one of the solutions to our equation. We integrate the last expression: ![]() where C is the constant of integration. Returning to the old variable y, we can write: ![]() Thus, the equation has two solutions: ![]() |
Example 2 |
Solve the differential equation Solution. Note that the root x = 0 does not belong to the domain of the given differential equation. Let us rewrite the equation in the following form: ![]() As can be seen, the equation is homogeneous. Let us make the substitution y = ux. Therefore, ![]() We substitute the resulting expression into the differential equation: ![]() Let’s divide both sides by x ≠ 0 : ![]() As a result, we obtain an equation with separable variables: ![]() In the next step, we integrate the left and right sides of the equation: ![]() Hence, ![]() The constant C here can be written as ln C 1 ( C 1 > 0). Then ![]() Thus, we got two solutions: ![]() If C 1 = 0, then the answer is the function y = xe. It is easy to verify that this function will also be the solution to the differential equation. Indeed, substituting ![]() into the differential equation, we find: ![]() Thus, all solutions of the differential equation can be represented by one formula: ![]() where C is an arbitrary real number. |
Example 3 |
Solve the differential equation Solution. Here we again encounter a homogeneous equation. Indeed, let’s write it as: ![]() Let’s make the substitution y = ux. Then y’ = u’x + u . Substituting y and y’ into the original equation, we get: ![]() Let’s divide both sides of the equation by ux 2. Note that the root x = 0 is not a solution, but we can verify that the root u = 0 (or y = 0 ) is one of the solutions of this differential equation. As a result, we obtain: ![]() By integrating, we find the general solution: ![]() Considering that ![]() The inverse function x ( y ) has an explicit form: ![]() Since C is an arbitrary number, the minus sign in front of this constant can be replaced with a plus sign. Then we get: ![]() Thus, the differential equation has solutions: ![]() |
Example 4 |
Solve the differential equation Solution. From the form of the right-hand side of the equation, it follows that x ≠ 0 and y ≠ 0. We can make the substitution: y = ux, y’ = u’x + u, which leads to an equation with separable variables: ![]() Integrating this equation, we obtain: ![]() Let us redesignate 2 C simply as the constant C. Therefore, ![]() So, the general solution is written as: ![]() |
Example 5 |
Find the general solution of the differential equation Solution. As can be seen, this equation is homogeneous. Therefore, we use the substitution y = ux, y’ = u’x + u. As a result, the equation is transformed into an equation with separable variables: ![]() Divide both sides by x 3. (Note that the root x = 0 is not a solution.) ![]() Now we can integrate the last equation: ![]() Since u = y/x, the solution is written as: ![]() It follows that ![]() Let us re-designate for brevity: e C = C 1, ( C 1 > 0). Then the solution in implicit form is determined by the equation: ![]() where the constant C 1 > 0. |
Example 6 |
Solve the equation Solution. The numerator and denominator on the right-hand side correspond to intersecting lines. Therefore, this differential equation can be transformed into a homogeneous one by an appropriate coordinate transformation. Let the new and old coordinates be related by the following relations: ![]() We will determine the constants α and β later. Substituting these relations into the equation, we obtain: ![]() In the new coordinates, the differential equation takes the form: ![]() This equation will be homogeneous if the coefficients α and β satisfy the system of equations. ![]() Solving this system of equations for α and β, we find: ![]() For the given values of α and β, the differential equation is written as follows: ![]() We have obtained a homogeneous equation. Next, we make the substitution: Y = uX, where u is some function of X . Therefore, dY = Xdu + udX. As a result, we have: ![]() We divide the numerator and denominator on the right-hand side by X . We can verify that X = 0 or x = X + α = −1/2 is not a solution to the differential equation. Simple transformations lead to the following result: ![]() Let’s expand the quadratic function in the numerator of the fraction on the right-hand side into a product of monomials: ![]() Hence, ![]() Then ![]() By separating the variables, we can write: ![]() We integrate the resulting equation: ![]() Now let’s transform the integrand on the left-hand side. We’ll use the method of undetermined coefficients and expand the integrand into a sum of rational fractions: ![]() Hence, ![]() Thus, the differential equation is written as follows: ![]() After integrating both parts, we obtain: ![]() where the constant C is a positive real number. We rewrite the solution in terms of the variables X and Y : ![]() It is convenient to further denote: 5 lnC = lnC 1, where C 1 is an arbitrary positive number. Thus, the solution can be written as: ![]() Now we can return to the original variables x, y. Since ![]() then we get: ![]() The right side can be simplified again somewhat: ![]() Then the final general solution of the original differential equation is expressed by the following implicit formula: ![]() where the constant C 3 is any number not equal to zero. |
Example 7 |
Find the general solution of the differential equation Solution. It can be seen that the equations of the lines in the numerator and denominator on the right-hand side correspond to parallel lines. Therefore, we make the following change of variables: ![]() As a result, the differential equation takes the form: ![]() As you can see, we’ve obtained a simple equation with separable variables. Solving it, we find the answer: ![]() From the last expression, we can derive an explicit function y ( x ) : ![]() Thus, ![]() |
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