99 Percentile Qs Bank for JEE MainMathematicsDifferential Equations
The particular solution of the different equation d x d y = y(1+y y) x (x^2 e ) , y(1)=0
Options
- Ay y=x^2 x
- By^2 y= x
- Cy= ( e^2 e )(x-1)
- Dy=e^2 x
Correct answer
A. y y=x^2 x
Step-by-step solution
aligned & d x d y = y(1+y y) x ( x ^2 e ) , y(1)=0 & x (x^2 e ) d x= ( y+y y) d y & Let x^2 e=t & x d x= 1 2 e d t & 1 2 e t d t=- y+y y- d y d y y d y+C & 1 2 e [t t-t]=y y+c aligned aligned & x^2 e 2 e ( x^2 e-1 )=y y+c & x^2 2 ( x^2+ e-1 )=y y+C & x^2 x=y y+C _ e ^ e =1 & When x=1, y=0 & 1 1=0+C & C=0 aligned Hence x^2 x=y y