The differential equation $ \frac{dy}{dx} = \frac{3x + y + 3}{xy + 2x + 4y + 8} $ presents a classic challenge in introductory differential equations courses. At first glance, the mixture of $x$ and $y$ terms in both the numerator and denominator suggests a complex, perhaps non-separable structure. That said, a closer inspection of the algebraic architecture reveals a pathway to a solution through strategic factorization and variable separation. This article provides a comprehensive, step-by-step guide to solving this equation, explaining the mathematical reasoning behind each manipulation Practical, not theoretical..
Understanding the Structure of the Equation
The given Ordinary Differential Equation (ODE) is:
$ \frac{dy}{dx} = \frac{3x + y + 3}{xy + 2x + 4y + 8} $
This is a first-order, first-degree differential equation. The standard approach for such equations involves checking for separability, exactness, homogeneity, or the potential for an integrating factor.
Looking at the denominator $xy + 2x + 4y + 8$, we notice a grouping pattern. The terms can be rearranged by factoring by grouping:
$ xy + 2x + 4y + 8 = x(y + 2) + 4(y + 2) = (x + 4)(y + 2) $
This factorization is the critical breakthrough. It transforms the denominator from a sum of four terms into a product of two linear binomials Worth keeping that in mind..
Now, observe the numerator: $3x + y + 3$. But while it does not factor with the denominator directly, we can manipulate it to align with the factors $(x+4)$ and $(y+2)$. We want to express the numerator as a combination of these binomials.
Easier said than done, but still worth knowing.
$ A(x+4) + B(y+2) = Ax + 4A + By + 2B $
Matching coefficients with $3x + y + 3$:
- Coefficient of $x$: $A = 3$
- Coefficient of $y$: $B = 1$
- Constant term: $4A + 2B = 4(3) + 2(1) = 12 + 2 = 14$
The constant term matches ($14 \neq 3$). Practically speaking, the constant is 3. And numerator: $3x + y + 3$. On top of that, $4(3) + 2(1) = 14$. Consider this: wait, the constant in the numerator is 3. Let's re-check the problem statement: "dy dx xy 3x y 3 xy 2x 4y 8". Let's re-read the numerator: $3x + y + 3$. Also, there is a discrepancy. Denominator: $xy + 2x + 4y + 8 = (x+4)(y+2)$.
If the numerator were $3x + y + 14$, it would be $3(x+4) + 1(y+2)$. Since it is $3x + y + 3$, we have $3x + y + 3 = 3(x+4) + (y+2) - 11$. $3x + 12 + y + 2 - 11 = 3x + y + 3$. Correct The details matter here..
So the numerator is $3(x+4) + (y+2) - 11$.
The differential equation becomes: $ \frac{dy}{dx} = \frac{3(x+4) +
