Differential Equations
Differential Equations
nta_pyq_2025_jan
Grade 12
Question:
Let x = x(y) be the solution of the differential equation y 2 dx + (x - 1 ) dy = 0 . If x(1) = 1, then x ( 1 ) is : y 2
1 2 + e
3 + e
3 - e
3 2 + e
Step-by-Step Solution
Key Concept: Apply the core result for formation and solution of differential equations and simplify using the given constraints.
2 y dx + (x - 1 ) dy = 0 (3) y 1 2 y dx = ( - x) dy y dx 1 2 \Rightarrow y = - x dy y dx x 1 \Rightarrow + = 2 3 dy y y 1 \int dy -1 2 I.F. = e y = e y \therefore Solution is -1 1 - 1 y y xe = \int e \times dy + C 3 y Let -1 = t y 1 \Rightarrow dy = dt 2 y 1 - t \Rightarrow xe y = -\int e tdt + C 1 - t \Rightarrow xe y = -e (t - 1) + C - 1 -1 -1 y y \Rightarrow xe = -e ( - 1) + C y x(1) = 1 -1 -1 \Rightarrow e = -e (-2) + C -1 \Rightarrow C = -e 1 1 -1+ y \Rightarrow x = + 1 - e y 1 x( ) = 3 - e 2 -1
Correct Answer: 3