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) → $y^2 = x^2+2 \log_e|1+x|+1$
$(y-x^2y)dy=(1-x^3)dx$
$⇒y(1-x^2)dy=(1-x^3)dx$
$⇒ydy=\frac{1-x^3}{1-x^2}dx$
$⇒\int ydy=\int\frac{(1-x)(1+x^2+x)}{(1-x)(1+x)}dx$
$⇒\frac{y^2}{2}=\int\frac{x(1+x)+1}{(1+x)}dx$
$⇒\frac{y^2}{2}=\frac{x^2}{2}+\int\frac{1}{1+x}dx$
$⇒y=x^2+2\log|x+1|+2C$
and,
$y(0)=1$
$⇒1=0+0+2C$
$⇒C=\frac{1}{2}$
$∴y=x^2+2\log|x+1|+1$