Question
Solution of the equation xdy=(y+xf(yx)f′(yx))dx
Answer: Option B
:
B
We have, xdy=(y+xf(yx)f′(yx))dx
⇒dydx=yx+f(yx)f′(yx)which is homogenous
Putting y=vx⇒dydx=v+xdvdx, we obtain
v+xdvdx=v+f(v)f′(v)⇒f′(v)f(v)dv=dxx
Integrating, we get log f(v)=logx+logc
⇒logf(v)=logcx⇒f(yx)=cx
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:
B
We have, xdy=(y+xf(yx)f′(yx))dx
⇒dydx=yx+f(yx)f′(yx)which is homogenous
Putting y=vx⇒dydx=v+xdvdx, we obtain
v+xdvdx=v+f(v)f′(v)⇒f′(v)f(v)dv=dxx
Integrating, we get log f(v)=logx+logc
⇒logf(v)=logcx⇒f(yx)=cx
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