Solution of differential equation $(x-y)

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Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Differential Equations

Question:

Solution of differential equation $(x-y)^2\frac{dy}{dx}=1$ is:

Options:

$y=\frac{1}{2}\log|\frac{x-y-1}{x-y+1}|+c$

$y=\frac{1}{2}\log|\frac{x+y-1}{x+y+1}|+c$

Both A and B

None of these

Correct Answer:

$y=\frac{1}{2}\log|\frac{x-y-1}{x-y+1}|+c$

Explanation:

$x - y = t$ $⇒1-\frac{dy}{dx}=\frac{dt}{dx}⇒1-\frac{dt}{dx}=\frac{1}{t^2}⇒\int dx=\int\frac{t^2}{t^2-1}dx$

$⇒y=\frac{1}{2}\log|\frac{t-1}{t+1}|=c⇒y=\frac{1}{2}\log|\frac{x-y-1}{x-y+1}|+c$