What is the solution of the differential equation cos(dy/dx) = p when y(0) = q?
A.cos((y-q)/x) = p✓ Correct
B.cos((y-p)/x) = q
C.cos⁻¹((y-q)/x) = p
D.cos⁻¹((y-p)/x) = q
Explanation
From cos(dy/dx) = p, we get dy/dx = cos⁻¹(p) = constant. Integrating: y = x·cos⁻¹(p) + C. Using y(0) = q gives C = q, so y - q = x·cos⁻¹(p), meaning cos⁻¹(p) = (y-q)/x, i.e., cos((y-q)/x) = p.
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