Question

18. The general solution of the differential equation e* dy + (y e2x) dx 0 is(A) xexC(C) ye2C

Answer 1

The given differential equation is,

e x dy+( y e x +2x )dx=0 dy dx = −( y e x +2x ) e x dy dx = −y e x e x − 2x e x dy dx +y=− 2x e x

The given differential equation is of the form dy dx +Py=Q.

Here,

P=1 Q=− 2x e x

Now, calculate the Integrating factor,

IF= e ∫ Pdx = e ∫ 1dx = e x

The solution of the differential equation is,

y( IF )= ∫ ( Q×IF )dx +c y e x = ∫ ( − 2x e x e x ) dx+c y e x =−2 ∫ x dx+c y e x =−2 x 2 2 +c

Further, simplify the equation,

y e x =− x 2 +c y e x + x 2 =c

Hence, the correct option is ©.