A) \[{{e}^{2}}-1\]
B) \[{{e}^{2}}+1\]
C) \[2({{e}^{2}}-1)\]
D) \[-2({{e}^{2}}-1)\]
Correct Answer: C
Solution :
[c] : We have, \[\frac{dy}{dx}=(x+3)(y+2)\] Take, X=x+3 and \[Y=y+2\Rightarrow \frac{dY}{Y}=XdX\] \[Y=A{{e}^{{{X}^{2}}/2}}\Rightarrow y=-2+A{{e}^{\frac{{{(x+3)}^{2}}}{2}}}\] Now, \[y(-1)=-2+A{{e}^{2}},y(-3)=-2+A\] \[\therefore \]\[y(-1)-{{e}^{2}}y(-3)=-2+2{{e}^{2}}=2({{e}^{2}}-1)\].
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