Solve the following initial value differential equation
(x − 1) dy/dx = 2xy, when y(2) = 1.
This is a variable separable differential equation.
Separate variables
(x − 1) dy/dx = 2xy
dy/y = [2x / (x − 1)] dx
Integrate both sides
∫ dy/y = ∫ [2x / (x − 1)] dx
2x/(x−1) = 2 + 2/(x−1) using long division or manipulation of the numerator
∫ dy/y =∫ [2 + 2/(x − 1)] dx
ln|y| = 2x + 2ln|x − 1| + C
Apply initial condition y(2) = 1
When x = 2, y = 1
ln|1| = 2(2) + 2ln|2 − 1| + C
0 = 4 + 2ln(1) + C
0 = 4 + 0 + C
⇒ C = −4
ln|y| = 2x + 2ln|x − 1| − 4
ln|y| - 2ln|x − 1| = 2x − 4
using property of loagarithms
ln|y| - ln|x − 1|² =2x − 4
ln [|y| / |x − 1|² ] =2x − 4
y = (x − 1)² e^(2x − 4)
see this video for more explanation
cbse 12th applied mathematics variable separable differential equation previous year question papers 2025 2026
