Solving Homogeneous And Non-Homogeneous Differential Equations

In the realm of differential equations, understanding the concept of homogeneity is paramount. To properly tackle homogeneous equations, let's dive deep into the first question, which asks us to determine the degree of homogeneity for the differential equation ${2ye^x \, dx + (y - 2xe^y) \, dy = 0}$.

To determine if a differential equation is homogeneous, we must first rearrange it into the form ${\frac{dy}{dx} = f(x, y)}$. Given the equation ${2ye^x \, dx + (y - 2xe^y) \, dy = 0}$, we can rewrite it as:

${ (y - 2xe^y) \, dy = -2ye^x \, dx }$ ${ \frac{dy}{dx} = \frac{-2ye^x}{y - 2xe^y} }$

Now, for a function ${f(x, y)}$ to be homogeneous of degree ${n}$, it must satisfy the condition ${f(tx, ty) = t^n f(x, y)}$ for any scalar ${t}$. Let's apply this to our function:

${ f(x, y) = \frac{-2ye^x}{y - 2xe^y} }$

Substitute ${x}$ with ${tx}$ and ${y}$ with ${ty}$:

${ f(tx, ty) = \frac{-2(ty)e^{tx}}{ty - 2(tx)e^{ty}} }$

This expression does not neatly simplify into the form ${t^n f(x, y)}$. This suggests that the initial equation presented might be misleadingly phrased, as it does not fit the classical definition of a homogeneous function directly applicable to the degree concept. Instead, let's focus on the differential equation's structure to determine homogeneity in a broader sense. The equation is in the form ${M(x, y)dx + N(x, y)dy = 0}$, where ${M(x, y) = 2ye^x}$ and ${N(x, y) = y - 2xe^y}$. For the equation to be homogeneous, the functions ${M}$ and ${N}$ should be homogeneous of the same degree. However, due to the presence of the exponential terms, it's not homogeneous in the traditional polynomial sense where degrees can be directly compared by variable powers. The exponential functions ${e^x}$ and ${e^y}$ disrupt the homogeneity because substituting ${tx}$ and ${ty}$ does not yield a common power of ${t}$ that can be factored out.

Therefore, a more appropriate analysis involves recognizing that while the equation isn't homogeneous in the standard algebraic sense (where we sum the powers of variables), the terms' structure indicates it isn't easily classified by degree. The exponential components prevent a straightforward degree assignment, leading to the conclusion that the question's premise might be flawed in its direct applicability of a degree. This nuanced understanding is critical because misclassifying such equations can lead to incorrect solution approaches. Recognizing that this equation is not homogeneous in the typical polynomial sense helps in choosing the correct method to solve it, possibly involving techniques for non-homogeneous equations or special integrating factors. Therefore, the presence of exponential functions fundamentally changes how we assess homogeneity, making standard degree calculations inapplicable and emphasizing the importance of understanding various equation forms and solution strategies.

The second question presents us with the differential equation ${y'' + 3y' - 2y = 7x}$ and asks us to classify it. The heart of this classification lies in understanding the difference between homogeneous and non-homogeneous differential equations. In the context of linear differential equations, the distinction is quite clear.

A differential equation is termed homogeneous if, when set to zero, it satisfies the superposition principle. This means that if ${y_1(x)}$ and ${y_2(x)}$ are solutions to the homogeneous equation, then any linear combination ${c_1y_1(x) + c_2y_2(x)}$ is also a solution, where ${c_1}$ and ${c_2}$ are constants. Crucially, a homogeneous linear differential equation has the form:

${ a_n(x)y^{(n)} + a_{n-1}(x)y^{(n-1)} + \cdots + a_1(x)y' + a_0(x)y = 0 }$

Notice the key characteristic: the equation equals zero. This is what allows solutions to be combined linearly and still satisfy the equation. In contrast, a non-homogeneous differential equation has a non-zero function on the right-hand side:

${ a_n(x)y^{(n)} + a_{n-1}(x)y^{(n-1)} + \cdots + a_1(x)y' + a_0(x)y = g(x) }$

where ${g(x)}$ is a non-zero function. This presence of ${g(x)}$ fundamentally changes the nature of the solutions. The superposition principle no longer holds in the same way; linear combinations of solutions to the homogeneous part of the equation (i.e., the equation with ${g(x)}$ replaced by 0) are not necessarily solutions to the non-homogeneous equation.

Now, let's apply this knowledge to our specific equation, ${y'' + 3y' - 2y = 7x}$. Comparing this with the general forms above, we immediately see that it matches the structure of a non-homogeneous equation. The term ${7x}$ on the right-hand side is a non-zero function, which explicitly makes the equation non-homogeneous. The implication of this non-homogeneity is significant for solving the equation. The general solution to a non-homogeneous linear differential equation consists of two parts: the complementary solution and the particular solution. The complementary solution, ${y_c(x)}$, is the general solution to the associated homogeneous equation (obtained by setting the right-hand side to zero). In our case, this would be the solution to ${y'' + 3y' - 2y = 0}$. The particular solution, ${y_p(x)}$, is any specific solution to the non-homogeneous equation. The general solution to the non-homogeneous equation is then the sum of these two:

${ y(x) = y_c(x) + y_p(x) }$

Finding ${y_p(x)}$ often involves methods like undetermined coefficients or variation of parameters, which are specifically designed to tackle the non-homogeneous part of the equation. Understanding this structure is crucial for selecting the appropriate solution technique. For our equation, the method of undetermined coefficients would be a viable approach since ${7x}$ is a polynomial. We would assume a particular solution of the form ${Ax + B}$, where ${A}$ and ${B}$ are constants to be determined, and substitute this into the original equation. This understanding of classifying differential equations as homogeneous or non-homogeneous is not just an academic exercise; it's a practical tool that guides the entire solution process. Recognizing the non-homogeneous nature of the equation immediately directs us toward the appropriate methods for finding both the complementary and particular solutions, ensuring we arrive at the correct general solution.

Solving differential equations involves a range of methodologies, each tailored to specific types of equations. The equation ${2ye^x \, dx + (y - 2xe^y) \, dy = 0}$ and the equation ${y'' + 3y' - 2y = 7x}$ exemplify this diversity. Addressing the first equation, ${2ye^x \, dx + (y - 2xe^y) \, dy = 0}$, requires a careful examination to determine the most suitable approach. As previously discussed, this equation isn't homogeneous in the traditional polynomial sense. Therefore, standard methods for homogeneous equations, such as substituting ${y = vx}$, are not directly applicable. Instead, we should explore alternative strategies such as exact equations or integrating factors. To determine if the equation is exact, we check if:

${ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} }$

where ${M(x, y) = 2ye^x}$ and ${N(x, y) = y - 2xe^y}$. Let's compute the partial derivatives:

${ \frac{\partial M}{\partial y} = 2e^x }$

${ \frac{\partial N}{\partial x} = -2e^y }$

Since ${2e^x \neq -2e^y}$, the equation is not exact. This directs us toward finding an integrating factor. An integrating factor, ${\mu(x, y)}$, is a function that, when multiplied by the original equation, makes it exact. Finding an integrating factor can be complex, but sometimes, a function of only ${x}$ or ${y}$ will suffice. We can test for integrating factors dependent only on ${x}$ or ${y}$ using the following formulas:

For ${\mu(x)}$:

${ \frac{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}}{M} = f(x) }$

For ${\mu(y)}$:

${ \frac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{N} = g(y) }$

If either ${f(x)}$ or ${g(y)}$ is a function of only ${x}$ or ${y}$, respectively, then we can find the integrating factor. Let's try to find ${f(x)}$:

${ \frac{-2e^y - 2e^x}{2ye^x} = \frac{-(e^y + e^x)}{ye^x} }$

This is not a function of only ${x}$. Now, let's try to find ${g(y)}$:

${ \frac{2e^x - (-2e^y)}{y - 2xe^y} = \frac{2(e^x + e^y)}{y - 2xe^y} }$

This is also not a function of only ${y}$. The absence of a simple integrating factor suggests that a more advanced technique or numerical method may be required to solve this equation. The complexity arises from the mixture of exponential and polynomial terms, making standard solution techniques less effective.

Turning our attention to the second equation, ${y'' + 3y' - 2y = 7x}$, we recognize it as a linear, second-order, non-homogeneous differential equation with constant coefficients. As previously discussed, the general solution is the sum of the complementary solution ${y_c(x)}$ and the particular solution ${y_p(x)}$. To find ${y_c(x)}$, we solve the associated homogeneous equation:

${ y'' + 3y' - 2y = 0 }$

This involves finding the roots of the characteristic equation:

${ r^2 + 3r - 2 = 0 }$

Using the quadratic formula, we find the roots:

${ r = \frac{-3 \pm \sqrt{3^2 - 4(1)(-2)}}{2(1)} = \frac{-3 \pm \sqrt{17}}{2} }$

Thus, the complementary solution is:

${ y_c(x) = c_1e^{r_1x} + c_2e^{r_2x} }$

where ${r_1 = \frac{-3 + \sqrt{17}}{2}}$ and ${r_2 = \frac{-3 - \sqrt{17}}{2}}$, and ${c_1}$ and ${c_2}$ are arbitrary constants.

For the particular solution ${y_p(x)}$, we use the method of undetermined coefficients. Since the non-homogeneous term is ${7x}$, we assume a solution of the form:

${ y_p(x) = Ax + B }$

We find the first and second derivatives:

${ y_p'(x) = A }$

${ y_p''(x) = 0 }$

Substituting these into the original non-homogeneous equation gives:

${ 0 + 3A - 2(Ax + B) = 7x }$

${ 3A - 2Ax - 2B = 7x }$

Equating coefficients, we get:

${ -2A = 7 \implies A = -\frac{7}{2} }$

${ 3A - 2B = 0 \implies B = \frac{3A}{2} = -\frac{21}{4} }$

So, the particular solution is:

${ y_p(x) = -\frac{7}{2}x - \frac{21}{4} }$

The general solution to the non-homogeneous equation is then:

${ y(x) = c_1e^{\frac{-3 + \sqrt{17}}{2}x} + c_2e^{\frac{-3 - \sqrt{17}}{2}x} - \frac{7}{2}x - \frac{21}{4} }$

This comprehensive approach highlights the importance of correctly identifying the type of differential equation to apply the appropriate solution method. For the first equation, the complexity lies in finding a suitable integrating factor, while the second equation demonstrates the standard procedure for solving linear, second-order, non-homogeneous equations with constant coefficients. Each type demands a specific strategy, underscoring the need for a versatile toolkit in differential equation solving.

In summary, differential equations exhibit a diverse range of forms and require tailored solution methodologies. Recognizing the characteristics of equations, such as homogeneity, exactness, and linearity, is crucial for selecting the correct approach. The exploration of these two equations highlights the importance of a comprehensive understanding of differential equation theory and its practical applications in solving mathematical problems.