The solution of (x + log y)dy + ydx = 0

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Target Exam

CUET

Subject

-- Mathematics - Section A

Chapter

Differential Equations

Question:

The solution of (x + log y)dy + ydx = 0 when y(0) = 1 is:

Options:

y(x - 1) + y log y = 0

y(x - 1 + y log y) + 1 = 0

xy + y log - y + 1 = 0

None of these

Correct Answer:

xy + y log - y + 1 = 0

Explanation:

$\int(xdy+ydx)+\int\log ydy=0⇒\int d(xy)+\int\log y dy = c$

⇒ xy + y log y - y = c; y(0) - 1 ⇒ c = -1 ⇒ xy + y log y - y + 1 = 0