The particular solution of the differential equation \((y-x^2y)dy = (1-x^3)dx\) with \(y(0) = 1\), is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(y^2 = x^2+2\log_e|1+x|+1\)
The equation is variables separable. Taking y out on the left and factorising both sides:
\(y(1-x^2)\,dy = (1-x^3)\,dx\)
\( \displaystyle y\,dy = \frac{(1-x)(1+x+x^2)}{(1-x)(1+x)}\,dx \)
Cancelling 1 − x and writing \(1+x+x^2 = x(1+x)+1\):
\( \displaystyle y\,dy = \left(x+\frac{1}{1+x}\right)dx \)
Integrating both sides:
\( \displaystyle \frac{y^2}{2} = \frac{x^2}{2} + \log_e|1+x| + C \)
Putting x = 0, y = 1:
\( \displaystyle \frac{1}{2} = 0 + \log_e 1 + C,\quad C = \frac{1}{2} \)
Multiplying by 2:
\(y^2 = x^2 + 2\log_e|1+x| + 1\)
Why the other options are incorrect
Option (2) \(y^2=1+x^2+2\log_e\left|\frac{1+x}{2}\right|\) — Incorrect. At x = 0 it gives \(y^2 = 1 + 2\log_e\frac{1}{2} = 1 - 2\log_e 2\), not 1, so it fails y(0) = 1.
Option (3) \(y^2=x^2+2x-3\) — Incorrect. At x = 0 it gives \(y^2 = -3\), which is impossible.
Option (4) \(y^2 = x^2+2x+1\) — Incorrect. It does pass y(0) = 1, but differentiating gives \(y\,dy = (x+1)\,dx\), and \(x + 1 \ne x + \frac{1}{1+x}\), so it does not satisfy the differential equation.