Question

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

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Solution

## 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 ©.

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