Question

The solution of the differential equation `dy/dx=1+y/x+(y/x)^2` is ______.

विकल्प

• `tan^-1(y/x)=logx+c`

• `sin^-1(y/x)=-logx+c`

• `sin^-1(y/x)=logx+c`

• `tan^-1(y/x)=-logx+c`

Answer 1

The solution of the differential equation `dy/dx=1+y/x+(y/x)^2` is `underline(tan^-1(y/x)=logx+c)`.

Explanation:

`dy/dx=1+y/x+(y/x)^2`            ....(i)

Put `v=y/x=>y=vx`            ....(ii)

`therefore dy/dx=v+x(dv)/dx`        ....(iii)

Substituting (ii) and (iii) in (i), we get

`v+x(dv)/dx=1+v+v^2`

`=>x(dv)/dx=1+v^2=>(dv)/(1+v^2)=dx/x`

Integrating on both sides, we get

tan^-1 v = log x + c

`=>tan^-1(y/x)=logx+c`