Questions
a
y2e−x+2lny=c
b
y2e−2x−2lny=c
c
y2e−2x+2lny=c
d
y2e−x−2lny=c
detailed solution
Correct option is C
We have dxdy=y2+e2xy3⇒dxdy=1y+e2xy3⇒e−2xdxdy−e−2x1y=1y3→1Let −e−2x2=u→2⇒e2xdxdy=dudy→3Substituting 2,3 in 1, we getdudy+2yu=1y3whose solution is u.y2=∫1ydy+k⇒uy2=lny+k⇒−e−2x2y2=lny+k ∵u=−e−2x2⇒−e−2xy2=2lny+2k∴e−2xy2+2lny=c where c is an arbitrary constantTalk to our academic expert!
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