99 Percentile Qs Bank for JEE MainMathematicsDifferential Equations
The general solution of the differential equation ( d y d x +y g^ (x)=g(x) g^ (x) ) is
Options
- A(g(x)+ (1+y+g(x))=c )
- B(g(x)+ (1+y-g(x))=c )
- C(g(x)- (1+y+g(x))=c )
- D(g(x)- (1+y-g(x))=c )
Correct answer
B. (g(x)+ (1+y-g(x))=c )
Step-by-step solution
( aligned & d y d x +y g^ (x)=g(x) g^ (x) & P=g^ (x) and Q=g(x) g^ (x) & IF =e^ p d x & =e^ g^ (x) d x & IF =e^ g(x) aligned ) General Solution is ( aligned y( IF ) & = Q( IF ) d x+c y (e^ g(x) ) & = g(x) g^ (x) e^ g(x) d x+c aligned ) Put, (g(x)=t ) ( gathered g^ (x) d x=d t y e^ g(x) = t e^t d t+e^c y e^ g(x) =e^t(t-1)+e^c y e^ g(x) =e^t t-e^t+e^c y e^ g(x) =e^ g(x) g(x)-e^ g(x) +e^c e^ g(x) [y+1-g(x)]=e^c gathered ) Taking ( ) on both sides, ( aligned _e e^ g(x) [y+1-g(x)] & = _e e^c g(x)+[y+1-g(x)] & =C aligned