Target Exam

CUET

Subject

Section B1

Chapter

Differential Equations

Question:

Find the general solution of the differential equation $ye^{x/y} dx = (xe^{x/y} + y^2) dy, y \neq 0$.

Options:

$e^{x/y} = y + C$

$x = y \ln(y + C)$

$e^{x/y} - 2y = C$

$2e^{x/y} = y - C$

Correct Answer:

$e^{x/y} = y + C$

Explanation:

The correct answer is Option (1) → $e^{x/y} = y + C$ ##

Given differential equation can be written as

$\frac{dx}{dy} = \frac{xe^{x/y} + y^2}{ye^{x/y}}$

Put $\frac{x}{y} = v$

$\Rightarrow \frac{dx}{dy} = v + y \frac{dv}{dy}$

$v + y \frac{dv}{dy} = \frac{ve^v + y}{e^v}$

$\Rightarrow y \frac{dv}{dy} = \frac{y}{e^v}$

$∴\int e^v \, dv = \int dy$

$e^v = y + C$

$\Rightarrow e^{x/y} = y + C$, which is the required solution.