The particular solution of the differential equation $(y-x^2y)dy = (1-x^3)dx$ with $y(0) = 1$, is:

$y^2 = x^2+2 \log_e|1+x|+1$

The correct answer is Option (1) → $y^2 = x^2+2 \log_e|1+x|+1$

$(y-x^2y)dy=(1-x^3)dx$

$⇒y(1-x^2)dy=(1-x^3)dx$

$⇒ydy=\frac{1-x^3}{1-x^2}dx$

$⇒\int ydy=\int\frac{(1-x)(1+x^2+x)}{(1-x)(1+x)}dx$

$⇒\frac{y^2}{2}=\int\frac{x(1+x)+1}{(1+x)}dx$

$⇒\frac{y^2}{2}=\frac{x^2}{2}+\int\frac{1}{1+x}dx$

$⇒y=x^2+2\log|x+1|+2C$

and,

$y(0)=1$

$⇒1=0+0+2C$

$⇒C=\frac{1}{2}$

$∴y=x^2+2\log|x+1|+1$