Question

Solve the given homogeneous equation implicitly. $$ y^{\prime}=\frac{x+2 y}{2 x+y} $$

Short answer

Question: Find the general solution for the given homogeneous equation: $$y' = \frac{x + 2y}{2x + y}$$. Answer: The general solution for the given homogeneous equation is $$y(\frac{y}{x} + 1) = x^2 \cdot e^C.$$

Step-by-step solution

Check for Homogeneity

Notice that if we replace x by kx and y by ky, we get $$ \frac{k(x + 2y)}{2kx + ky} = \frac{x + 2y}{2x + y}. $$ Therefore, the given equation is homogeneous.

Apply Substitution

Since the equation is homogeneous, we can use the substitution method. Let's make the following substitution: $$ v = \frac{y}{x}, \text{ so } y = vx. $$ Differentiate \(y\) with respect to \(x\) to obtain the expression of \(y'\): $$ y^{\prime} = v + x \cdot \frac{dv}{dx}. $$

Replace in the Given Equation

Replace \(y^{\prime}\) and \(y\) in the given equation using the expressions derived in Step 2: $$ v + x \cdot \frac{dv}{dx} = \frac{x + 2(vx)}{2x + vx}. $$

Simplify the Equation

Simplify the equation and separate the variables: $$ v + x \cdot \frac{dv}{dx} = \frac{3xv}{x+2v}. $$ Now, divide both sides by \(x\): $$ \frac{dv}{dx} = \frac{3v}{1+2v} - \frac{v}{x}. $$

Separate the Variables

Rewrite the equation in a separated form: $$ \frac{dv}{\frac{3v}{1+2v} - v} = \frac{dx}{x}. $$

Integrate Both Sides

Integrate both sides with respect to their respective variables: $$ \int \frac{1+2v}{v(1+v)}dv = \int \frac{1}{x} dx. $$

Solve the Integrals

Solve the integrals. For the left side, it is a partial fraction decomposition: $$ \frac{1+2v}{v(1+v)} = \frac{A}{v} + \frac{B}{1+v} $$ Consequently, we get \(A = 1\) and \(B = 1\). Now the equation is: $$ \int \left(\frac{1}{v} + \frac{1}{1+v}\right)dv = \int \frac{1}{x} dx. $$ Integrate both sides: $$ \ln |v| + \ln |v + 1| = \ln |x| + C. $$

Solve for \(v\)

Solve for \(v\) to find the general solution: $$ \ln |v(v + 1)| = \ln |x| + C. $$ Take exponent for both sides: $$ v(v + 1) = x \cdot e^C. $$

Express the Solution in Terms of \(y\) and \(x\)

Replace \(v\) by \(\frac{y}{x}\) to find the general solution in terms of \(y\) and \(x\): $$ \frac{y}{x}\left(\frac{y}{x} + 1\right) = x \cdot e^C. $$ Simplify the equation to get the general solution: $$ y(\frac{y}{x} + 1) = x^2 \cdot e^C, $$ which is the general solution for the given homogeneous equation.

Key concepts

Separation of Variables

Separation of variables is a useful technique for solving differential equations by isolating each variable on opposite sides of the equation. The goal is to express the differential equation in a form where all the terms involving one kind of variable are on one side, and terms involving the other kind are on the side with their respective differentials.
This often involves rewriting the differential equation, and then rearranging to separate the variables by performing algebraic manipulation. Once separated, each side can be integrated independently. This integration results in a solution linked through an arbitrary constant, typically represented as 'C'.
In this exercise, after simplifying the equation, we rewrite it to isolate the derivatives, which allows us to proceed to integration. It's crucial to correctly rearrange terms and carefully integrate both sides to solve for the original variables.

Partial Fraction Decomposition

Partial fraction decomposition is a method used to break down complex rational expressions into simpler fractions that are more easily integrable. We mainly use this method in integration when we encounter rational expressions for which direct integration is not straightforward.
In our example, after separating variables, we reached an expression that needed integration. The fraction on one side of the equation was simplified using partial fraction decomposition. The original expression, \( \frac{1+2v}{v(1+v)} \), is decomposed into parts, making it integrable. Here, we find coefficients 'A' and 'B' for each fraction, resulting in \( \frac{1}{v} \) and \( \frac{1}{1+v} \).
The key is setting up equations based on matching coefficients for all terms when equating to the original expression, which allows each part to be integrated separately.

Substitution Method

The substitution method is a pivotal technique when dealing with homogeneous differential equations. This method simplifies differential equations by reducing the number of variables involved, making them easier to solve.
In our homogeneous equation, we perform a substitution where \( v = \frac{y}{x} \) and consequently express \( y \) as \( vx \). This transforms the original equation in terms of the new variable \( v \), thereby simplifying the problem.
This technique leverages the homogeneity property of the equation, resulting in a new form suitable for algebraic manipulation, separation of variables, and subsequent integration. Finally, the substitution is reversed, transitioning back to the original variables, \( y \) and \( x \), to express the general solution.

Implicit Solution

An implicit solution for a differential equation represents a solution in a form where the dependent and independent variables appear intertwined, without an explicit expression for one variable solely in terms of the other.
In our example, after completing the integration, we arrive at the equation, \( v(v + 1) = x \cdot e^C \). Here, the function is expressed implicitly rather than solved explicitly for one variable. We may need additional methods like algebraic manipulation or initial conditions, depending on context, to express the solution explicitly.
It's important to understand that implicit solutions can still fully describe the behavior and relationship between the variables, even if they aren't isolated as a simple function of one in terms of the other.