Question

The general solution of the differential equation $\frac{d y}{d x}=\frac{x+y+1}{x+y-1}$ is given by

• A. $y = x \log (x+y) + c$
• B. $x - y = \log (x+y) + c$
• C. $x + y = \log (x+y) + c$
• D. $y = x + \log (x+y) + c$

Hint:

Whenever a differential equation contains a linear group like $(x+y)$ bundled together, that group is always the perfect substitution candidate. Splitting fractions after substitution makes the integration a straightforward step!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The problem asks for the general solution of a first-order non-homogeneous differential equation where the variables are combined in linear combinations of the form $(x+y)$.

Step 2: Key Formula or Approach:
Since the term $(x+y)$ is repeated in both the numerator and denominator, we can use a substitution method to convert the equation into a separable variable form: $$ \text{Put } x + y = u \implies 1 + \frac{d y}{d x} = \frac{d u}{d x} \implies \frac{d y}{d x} = \frac{d u}{d x} - 1 $$

Step 3: Detailed Explanation:
Substituting $x+y = u$ and replacing the derivative term into the original differential equation gives: $$ \frac{d u}{d x} - 1 = \frac{u + 1}{u - 1} $$ Add 1 to both sides to isolate the derivative term: $$ \frac{d u}{d x} = \frac{u + 1}{u - 1} + 1 = \frac{u + 1 + (u - 1)}{u - 1} = \frac{2u}{u - 1} $$ Now, separate the variables by shifting all $u$ terms to the left and $x$ terms to the right: $$ \left(\frac{u - 1}{2u}\right) d u = d x \implies \left(\frac{u - 1}{u}\right) d u = 2 d x $$ Split the left-hand fraction into two separate terms: $$ \left(1 - \frac{1}{u}\right) d u = 2 d x $$ Integrate both sides with respect to their corresponding variables: $$ \int \left(1 - \frac{1}{u}\right) d u = \int 2 d x $$ $$ u - \log|u| = 2x + c $$ Substitute back our original expression $u = x + y$: $$ (x + y) - \log(x + y) = 2x + c $$ Rearranging terms to isolate $y$ on the left side: $$ y = 2x - x + \log(x + y) + c $$ $$ y = x + \log(x + y) + c $$

Step 4: Final Answer:
The general solution of the differential equation is $y = x + \log (x+y) + c$, which corresponds to option (D).