Differential Equations
Differential Equations
Allen Star Batch
Grade 12
Question:
Solution of the differential equation $\left\{1 - \frac{y^2}{x(x-y)^2}\right\}dx + \left\{\frac{x^2}{(x-y)^2} - \frac{1}{y}\right\}dy = 0$ is :
$ln\left|\frac{x}{y}\right| + \frac{xy}{x-y} = c$
$\frac{xy}{x-y} = ce^{y/y}$
$ln|xy| = c + \frac{xy}{x-y}$
None of these
Step-by-Step Solution
Key Concept: Recognize that the differential equation can be decomposed into $\left(\frac{dx}{x} - \frac{dy}{y}\right) + \frac{x^2dy - y^2dx}{(x-y)^2} = 0$, where the second term equals $d\left(\frac{xy}{x-y}\right)$ upon careful verification of the differential.
Rewrite the equation as $\left(\frac{dx}{x} - \frac{dy}{y}\right) + \frac{x^2dy - y^2dx}{(x-y)^2} = 0$. This can be expressed as $\left(\frac{dx}{x} - \frac{dy}{y}\right) + \frac{d\left(\frac{y}{x}\right)}{\left(1 - \frac{1}{x/y}\right)^2} = 0$. Integrating gives $\ln|x| - \ln|y| - \frac{1}{\left(\frac{1}{x} - \frac{1}{y}\right)} = c$ or $\ln\left|\frac{x}{y}\right| + \frac{xy}{x-y} = c$.
Correct Answer: 1