Question

4. The solution of the differential equation
(1 Point)
(D - 2)?y = 0 is
cet + che + czet
ce2 + c2e2x + cze 2
(cı + C2x + C3x?) et
(C1 + C2x + C3x?) e2​

Answer 1

Given a differential equation of a conic.

(1+y2)dx=xydx⇒xdx=1+y2ydy

Integrating both the sides we get 

⇒lnx=21ln(1+y2)+lnC where lnC is an integration constant.

⇒2lnx=ln(1+y2)+2lnC⇒lnx2=ln(1+y2)+lnC′ where C′=C2

⇒x2=C′(1+y2)

Putting x=1,y=0 we get C′=1

∴ equation of C is x2=1+y2⇒x2−y2=1 (hyperbola)

eccentricity of the above conic r=e=1+

Answer 2

Answer:

u5w57s6tdt7d7tg8yg9ubugy8f8yft7xodaoOOsp