Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Differential Equations

Question:

Solution of (x + y - 1)dx + (2x + 2y - 3)dy = 0 is:

Options:

y + x + log (x + y - 2) = c

y + 2x + log (x + y - 2) = c

2y + x + log (x + y - 2) = c

2y + 2x + log (x + y - 2) = c

Correct Answer:

2y + x + log (x + y - 2) = c

Explanation:

$\frac{dy}{dx}=-\frac{x+y-1}{2x+2y-3}$

Put x + y = t;

$1+\frac{dy}{dx}=\frac{dt}{dx};\frac{dt}{dx}-1=-\frac{t-1}{2t-3}$

$\frac{dt}{dx}=\frac{t-2}{2t-3};\int\frac{(2t-3)dt}{t-2}=\int dx$;

2t + loge (t - 2) = x + c; x + 2y + loge (x + y - 2) = c