Question

Solve each differential equation by making a suitable transformation. $$ (5 x+2 y+1) d x+(2 x+y+1) d y=0 $$

Short answer

To solve the given differential equation \((5x + 2y + 1) dx + (2x + y + 1) dy = 0\), we first rewrite it as an ODE in the form \(\frac{dy}{dx} = P(x, y)\), where \(P(x, y) = \frac{2x + y + 1}{5x + 2y + 1}\). Then, we apply the transformation \(u = 5x + 2y\) and rewrite the ODE in terms of u. We obtain the transformed ODE as \(\frac{dy}{du} \cdot (5 + 2 \cdot \frac{dy}{dx}) = \frac{u - 2y + 1}{u}\). After separating variables and integrating both sides, we find the function \(y(u)\). Finally, we revert the transformation to obtain the function \(y(x)\), which is the solution to the given differential equation.

Step-by-step solution

Examine the given differential equation.

The given differential equation is: \[ (5x + 2y + 1) dx + (2x + y + 1) dy = 0 \] We are supposed to solve this in order to find the function \(y(x)\) that satisfies the above equation.

Convert the equation to a standard differential equation form.

In order to solve this differential equation, we will first rewrite it in the form of separation of variables. Let's divide by \(dx\) on both sides: \[ 5x + 2y + 1 + \frac{2x + y + 1}{dy/dx} = 0 \] Let's represent \(\frac{dy}{dx}\) as P(x,y): \[ P(x, y) = \frac{2x + y + 1}{5x + 2y + 1} \] Now we have a first-order ordinary differential equation (ODE) in the standard form: \[ \frac{dy}{dx} = P(x, y) \]

Find a suitable transformation.

To find a suitable transformation, let's examine the given equation. Since we have both x and y terms with similar coefficients (5, 2), let's try the substitution: \[ u = 5x + 2y \] We will also find the partial derivatives of \(u\) with respect to \(x\) and \(y\): \[ \frac{\partial u}{\partial x} = 5 \quad\text{and}\quad \frac{\partial u}{\partial y} = 2 \] Now, we need to change the original ODE to represent it in the form of \(u\): \[ \frac{dy}{dx} = \frac{2x + y + 1}{u} \] To express \(dy/dx\) in terms of u, we have: \[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \]

Rewrite the ODE using the transformation.

Now let's rewrite the ODE using the transformation. First, let's find \(\frac{du}{dx}\): \[ \frac{du}{dx} = 5 + 2 \cdot \frac{dy}{dx} \] And now, we substitute this back into the ODE: \[ \frac{dy}{du} \cdot (5 + 2 \cdot \frac{dy}{dx}) = \frac{2x + y + 1}{u} \] Since \(u = 5x + 2y\), we have: \[ \frac{dy}{du} \cdot (5 + 2 \cdot \frac{dy}{dx}) = \frac{u - 2y + 1}{u} \]

Solve the transformed ODE.

The transformed ODE is now simpler and we can solve it by separation of variables. Separate the variables: \[ \frac{dy}{du} = \frac{u - 2y + 1}{5u + 2u \cdot \frac{dy}{du}} \] Now, we can integrate both sides with respect to \(u\). After the integration, we can find the function \(y(u)\). Once we have found \(y(u)\), we can convert it back to the function of \(x\) using the transformation we applied earlier. Finally, the function \(y(x)\) will be the solution to the given differential equation.

Key concepts

Transformation Method

The transformation method is a powerful tool in solving differential equations, particularly when dealing with a complex ODE that might not be easily solvable using standard methods. In essence, it involves making a judicious substitution to transform the original equation into a more manageable form.

For example, in our exercise, the substitution \( u = 5x + 2y \) simplifies the given ODE. This is due to the identification of a pattern in the coefficients in the original ODE, which suggests a transformation that would make the equation separable. The ultimate goal is to rewrite the ODE in terms of a new variable \( u \) that will allow for easier manipulation and solution.

Once the new variable is introduced, the challenge becomes expressing the derivative of \( y \) with respect to \( x \) in terms of \( u \) as well. In doing so, one must find the partial derivatives of the new variable with respect to the old variables, \( x \) and \( y \) (\( \frac{\partial u}{\partial x} \) and \( \frac{\partial u}{\partial y} \) respectively), and then utilize them to convert the ODE fully into terms of \( u \). This will often lead to a more straightforward equation that one can solve using standard techniques.

Separation of Variables

Separation of variables is a technique used to solve a specific type of ODE that can be manipulated, so all terms containing one variable (for example, \( x \) or \( u \) in transformations) are on one side of the equation and all terms containing the other variable (such as \( y \)) are on the other side.

In our case, after substituting and rewriting the ODE with the new variable \( u \), we obtained a differential equation in \( u \) and \( y \) that could be separated \( \frac{dy}{du} = \frac{u - 2y + 1}{5u + 2u \cdot \frac{dy}{du}} \). The next step involves the careful rearrangement of the terms involving \( dy \) and \( du \) to opposite sides of the equation. After that, integration is applied to both sides, which will yield a relationship between \( u \) and \( y \)—in essence, solving for \( y(u) \).

Separation of variables is usually one of the first methods taught in differential equations since it applies to a variety of problems and if applicable, it provides a straightforward path to the solution.

First-Order Ordinary Differential Equation

A first-order ordinary differential equation (ODE) is an equation that involves the derivatives of a function but not its higher derivatives — meaning it contains only the first derivative \( dy/dx \). They often take the form \( \frac{dy}{dx} = f(x, y) \), where \( f \) is a given function of \( x \) and \( y \) that describes the rate of change of \( y \) with respect to \( x \).

The given exercise presents a first-order ODE, which we initially converted into a form that made P(x,y) clear: \( \frac{dy}{dx} = P(x, y) \). The solution to a first-order ODE is a function \( y(x) \) that satisfies this relationship between \( x \) and its derivative \( dy/dx \). Such equations are foundational in many fields, from physics to finance, because they can model a wide range of dynamic behaviors.

In practice, solving a first-order ODE often requires ingenuity in selecting an appropriate method. The separation of variables method used in our transformed equation is ideal for first-order ODEs when the equation can be manipulated into a separable form, thus allowing for integration and ultimately, finding the solution for \( y \) as a function of \( x \) or in our transformed case, as a function of \( u \) that can then be translated back to a function of \( x \) using the inverse transformation.