Question

Solve each differential equation by making a suitable transformation. $$ (10 x-4 y+12) d x-(x+5 y+3) d y=0. $$

Short answer

The solution to the given differential equation using the transformation \(v = x - y\) is: \[ \frac{1}{10}\ln|10(x - y) + 6y + 12| = C. \]

Step-by-step solution

Identify the transformation

We want to eliminate one of the variables in the differential equation. Let's choose the transformation v = x - y, based on the coefficients of the given equation. This will eliminate the x-terms or the y-terms, simplifying the equation.

Find the differentials

Now, we will find the differentials of v with respect to x and y. The transformation v = x - y has the following differentials: \[ dv = dx - dy \]

Rewrite the equation using the transformation

Substituting the differentials and the transformation v into the original equation, we get: \[ (10(v+y) - 4y + 12)(dv + dy) - (v + y + 5y + 3)dy = 0 \] Expanding the equation and simplifying, we obtain: \[ (10v + 6y + 12)dv = 0 \]

Separate the variables

Now that we've simplified the equation using the transformation v, we can separate the variables. Divide both sides of the equation by (10v + 6y + 12) to get: \[ \frac{dv}{10v + 6y + 12} = 0 \]

Integrate both sides

Next, we need to integrate both sides of the equation with respect to v and y, respectively: \[ \int \frac{dv}{10v + 6y + 12} = \int 0 dy \] Integrating the left-hand side with respect to v, we get: \[ \frac{1}{10}\ln|10v + 6y + 12| = C \] where C is the integration constant.

Solve for the original variables

Finally, we need to replace the transformation v = x - y back into the equation and solve for the original variables: \[ \frac{1}{10}\ln|10(x - y) + 6y + 12| = C \] This is the final solution to the given differential equation in terms of the original variables x and y.

Key concepts

Differential Equation Transformation

Understanding differential equation transformation is essential for simplifying complex equations. In essence, a transformation is a technique that allows us to convert a given differential equation into a more manageable form by introducing new variables. For example, in the provided exercise, the transformation chosen is v = x - y, which cleverly eliminates one set of terms, making the equation simpler to work with.

Transformations rely heavily on the properties of the equation at hand. The goal is to reduce the number of variables or to convert the equation into a recognizable form for which standardized solution methods apply. The differential for the new variable, in this case, is dv = dx - dy, showing how the differentials of x and y relate to the transformed variable v. This step is critical as it lays the groundwork for applying other solution methods, like the separation of variables.

Separation of Variables Method

The separation of variables method is a straightforward technique applied to differential equations where the variables can be isolated on opposite sides of the equation. This method, as seen in step 4 of the provided solution, involves dividing both sides of the simplified equation by an expression to isolate the differentials dv and dy.

Once the variables are separated, the problem reduces to integrating each side with respect to its variable. For our example, the equation simplifies to dv/(10v + 6y + 12) = 0, which indicates that the integrals of both sides with respect to their respective variables will lead us to the solution. This method is particularly powerful for first-order differential equations, which readily allow such separation.

Integrating Differential Equations

Integrating differential equations is the process of finding the antiderivative or integral of both sides of an equation to obtain the solution. In our exercise, the integration step comes after successfully employing the separation of variables method. The integration of dv/(10v + 6y + 12) = 0 simplifies the problem to calculating the integral of a single-variable function.

The outcome of this integration step is the antiderivative expressed with an arbitrary constant C, given that indefinite integration always includes an undetermined constant. The result, (1/10)ln|10v + 6y + 12| + C, represents a family of functions that solve the original differential equation. Determining the specific solution requires additional information, such as an initial condition or boundary value.

Differential Equation Solution

The differential equation solution we are after is a function or set of functions that satisfies the original problem. As evidenced in the given exercise, after integrating and applying antiderivatives, we arrive at a general solution that represents an infinite number of possible solutions, each corresponding to a different value of the constant C.

To finalize, we revert to the original variables x and y by undoing the transformation. The equation (1/10)ln|10(x - y) + 6y + 12| + C represents the general solution to the initial differential equation in terms of the original variables. It is these solutions that give us insight into the behavior of the system described by the differential equation, which can represent phenomena ranging from the motion of heavenly bodies to the flow of heat in a solid.