Explanation
y2dx+(x2−xy+42)dy=0
dxdy=−(x2−xy+y2)y2
dxdy=−x2y2+xy−1
Put v=x/y⟹x=vy
dx/dy=v+ydv/dy
⇒v+ydv/dy=−v2+v−1
⇒ydv/dy=−(v2+1)
dvv2+1=−1ydy
Integrating both side
⇒tan−1v=−logy+c
⇒tan−1x/y=−logy+cc=−c
⇒tan−1(x/y)+logy+c=0
dydx=gn+tan(yx)Itishomogenouous diffenrential equationLety=√x⇒dydx=√+xdvdxandputindifferentialequation⇒√+xdvdx=√+tanv⇒xdvdx=tanv⇒∫dvtanv=∫dxx⇒∫secv=logx+c⇒log|secv=tanv|logxc⇒log|secyx+tanyx|=logxc|secyx+tanyx|=xcAns.
Consider the given integral.
I=∫(xx)x(2xlogx+x)dx
I=∫(xx2)(2xlogx+x)dx
Let t=xx2
logt=x2logx
1tdtdx=(2xlogx+x)
dt=xx2(2xlogx+x)dx
Therefore,
I=∫1dt
I=t+C
On putting the value of t, we get
I=xx2+C
I=xxx+C
Hence, this is the answer.
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