PYQs / GATE CE / 2015 / Set 1 / Q48 GATE CE 2015 Set 1 — Question 48 Go beyond PYQs with Success Tracker AI-powered personalised practice and doubt support. Unlimited practice on eligible plans; AI usage limits apply. MCQ +2 / -0.67 Medium Exact Differential Equations Ordinary Differential Equations Engineering Mathematics
Engineering Mathematics → Ordinary Differential Equations → Exact Differential Equations
Last updated 5 September 2026
Question Consider the following differential equation:
x ( y d x + x d y ) cos y x = y ( x d y − y d x ) sin y x x(ydx + xdy) \cos \frac{y}{x} = y(xdy - ydx) \sin \frac{y}{x} x ( y d x + x d y ) cos x y = y ( x d y − y d x ) sin x y Which of the following is the solution of the above equation (
c c c is an arbitrary constant)?
Correct answer (C) xy cos (y)/(x) = c
Solution The given differential equation is:
x ( y d x + x d y ) cos y x = y ( x d y − y d x ) sin y x x(ydx + xdy) \cos \frac{y}{x} = y(xdy - ydx) \sin \frac{y}{x} x ( y d x + x d y ) cos x y = y ( x d y − y d x ) sin x y We can rewrite this using the differentials
d ( x y ) = x d y + y d x d(xy) = xdy + ydx d ( x y ) = x d y + y d x and
d ( y x ) = x d y − y d x x 2 d\left(\frac{y}{x}\right) = \frac{xdy - ydx}{x^2} d ( x y ) = x 2 x d y − y d x :
x d ( x y ) cos y x = y ⋅ x 2 d ( y x ) sin y x x d(xy) \cos \frac{y}{x} = y \cdot x^2 d\left(\frac{y}{x}\right) \sin \frac{y}{x} x d ( x y ) cos x y = y ⋅ x 2 d ( x y ) sin x y Dividing both sides by
x y ⋅ x cos y x xy \cdot x \cos \frac{y}{x} x y ⋅ x cos x y :
d ( x y ) x y = y ⋅ x 2 x y ⋅ x sin ( y / x ) cos ( y / x ) d ( y x ) \frac{d(xy)}{xy} = \frac{y \cdot x^2}{xy \cdot x} \frac{\sin(y/x)}{\cos(y/x)} d\left(\frac{y}{x}\right) x y d ( x y ) = x y ⋅ x y ⋅ x 2 cos ( y / x ) sin ( y / x ) d ( x y ) d ( x y ) x y = tan ( y x ) d ( y x ) \frac{d(xy)}{xy} = \tan\left(\frac{y}{x}\right) d\left(\frac{y}{x}\right) x y d ( x y ) = tan ( x y ) d ( x y ) Integrating both sides:
∫ d ( x y ) x y = ∫ tan ( y x ) d ( y x ) \int \frac{d(xy)}{xy} = \int \tan\left(\frac{y}{x}\right) d\left(\frac{y}{x}\right) ∫ x y d ( x y ) = ∫ tan ( x y ) d ( x y ) ln ( x y ) = ln ∣ sec ( y x ) ∣ + ln c \ln(xy) = \ln\left|\sec\left(\frac{y}{x}\right)\right| + \ln c ln ( x y ) = ln sec ( x y ) + ln c ln ( x y ) = ln ∣ c sec ( y x ) ∣ \ln(xy) = \ln\left|c \sec\left(\frac{y}{x}\right)\right| ln ( x y ) = ln c sec ( x y ) x y = c sec ( y x ) xy = c \sec\left(\frac{y}{x}\right) x y = c sec ( x y ) x y cos ( y x ) = c xy \cos\left(\frac{y}{x}\right) = c x y cos ( x y ) = c Thus, the solution is
x y cos y x = c xy \cos \frac{y}{x} = c x y cos x y = c , which corresponds to option (C).
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Correct answer (C) xy cos (y)/(x) = c
Solution The given differential equation is:
x ( y d x + x d y ) cos y x = y ( x d y − y d x ) sin y x x(ydx + xdy) \cos \frac{y}{x} = y(xdy - ydx) \sin \frac{y}{x} x ( y d x + x d y ) cos x y = y ( x d y − y d x ) sin x y We can rewrite this using the differentials
d ( x y ) = x d y + y d x d(xy) = xdy + ydx d ( x y ) = x d y + y d x and
d ( y x ) = x d y − y d x x 2 d\left(\frac{y}{x}\right) = \frac{xdy - ydx}{x^2} d ( x y ) = x 2 x d y − y d x :
x d ( x y ) cos y x = y ⋅ x 2 d ( y x ) sin y x x d(xy) \cos \frac{y}{x} = y \cdot x^2 d\left(\frac{y}{x}\right) \sin \frac{y}{x} x d ( x y ) cos x y = y ⋅ x 2 d ( x y ) sin x y Dividing both sides by
x y ⋅ x cos y x xy \cdot x \cos \frac{y}{x} x y ⋅ x cos x y :
d ( x y ) x y = y ⋅ x 2 x y ⋅ x sin ( y / x ) cos ( y / x ) d ( y x ) \frac{d(xy)}{xy} = \frac{y \cdot x^2}{xy \cdot x} \frac{\sin(y/x)}{\cos(y/x)} d\left(\frac{y}{x}\right) x y d ( x y ) = x y ⋅ x y ⋅ x 2 cos ( y / x ) sin ( y / x ) d ( x y ) d ( x y ) x y = tan ( y x ) d ( y x ) \frac{d(xy)}{xy} = \tan\left(\frac{y}{x}\right) d\left(\frac{y}{x}\right) x y d ( x y ) = tan ( x y ) d ( x y ) Integrating both sides:
∫ d ( x y ) x y = ∫ tan ( y x ) d ( y x ) \int \frac{d(xy)}{xy} = \int \tan\left(\frac{y}{x}\right) d\left(\frac{y}{x}\right) ∫ x y d ( x y ) = ∫ tan ( x y ) d ( x y ) ln ( x y ) = ln ∣ sec ( y x ) ∣ + ln c \ln(xy) = \ln\left|\sec\left(\frac{y}{x}\right)\right| + \ln c ln ( x y ) = ln sec ( x y ) + ln c ln ( x y ) = ln ∣ c sec ( y x ) ∣ \ln(xy) = \ln\left|c \sec\left(\frac{y}{x}\right)\right| ln ( x y ) = ln c sec ( x y ) x y = c sec ( y x ) xy = c \sec\left(\frac{y}{x}\right) x y = c sec ( x y ) x y cos ( y x ) = c xy \cos\left(\frac{y}{x}\right) = c x y cos ( x y ) = c Thus, the solution is
x y cos y x = c xy \cos \frac{y}{x} = c x y cos x y = c , which corresponds to option (C).
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