Question
The solution of dydx+xy=x2 is
Answer: Option B
:
B
dxdy+xy=x2⇒x−2dxdy+x−1y=1
Put x−1=t.Then−x−2dxdy=dtdy⇒x−2dxdy=−dtdy∴−dtdy+ty=1⇒dtdy−1yt=−1
It is linear in t. Here P=−1y.Q=−1
I.F=e∫(1y)dy=e−logy=1y
∴ Solution is t(1y)=∫(−1)1ydy=−logy+c⇒t=−ylogy+cy⇒x−1cy−ylogy.
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:
B
dxdy+xy=x2⇒x−2dxdy+x−1y=1
Put x−1=t.Then−x−2dxdy=dtdy⇒x−2dxdy=−dtdy∴−dtdy+ty=1⇒dtdy−1yt=−1
It is linear in t. Here P=−1y.Q=−1
I.F=e∫(1y)dy=e−logy=1y
∴ Solution is t(1y)=∫(−1)1ydy=−logy+c⇒t=−ylogy+cy⇒x−1cy−ylogy.
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