Q.

The general solution of the D.E y1+yφ1(x)−φ(x)φ1(x)=0 where φ(x) is a known function in

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a

y=ce-φ(x)+φ(x)-1

b

y=ceφ(x)+φ(x)-1

c

y=ce-φ(x)-φ(x)+1

d

y=ce-φ(x)+φ(x)+1

answer is A.

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Detailed Solution

dydx+yφ1(x)=φ(x)φ1(x)I⋅F=e∫φ′(x)dx=eφ(x) Solution of the D.E. yeφ(x)=∫eφ(x)⋅φ(x)φ1(x)dx yeφ(x)=∫et⋅tdt=et(t−1)+c yeφ(x)= eφ(x)φ(x)-1+cy=ce−φ(x)+φ(x)−1
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The general solution of the D.E y1+yφ1(x)−φ(x)φ1(x)=0 where φ(x) is a known function in