The general solution of the differential equation \( ye^{x/y}dx=\left(xe^{x/y}+y^2\right)dy \), \( (y\ne 0) \) is
(Where c is an arbitrary constant)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( e^{x/y}=y+c \)
The expression \( \frac{x}{y} \) suggests the substitution \( x=vy \), so that
\( \displaystyle dx=v\,dy+y\,dv \)
Substituting:
\( \displaystyle ye^{v}(v\,dy+y\,dv)=\left(vye^{v}+y^2\right)dy \)
\( \displaystyle vye^{v}\,dy+y^2e^{v}\,dv=vye^{v}\,dy+y^2\,dy \)
The \( vye^{v}\,dy \) terms cancel. Dividing by \( y^2\ (y\ne 0) \):
\( \displaystyle e^{v}\,dv=dy \)
\( \displaystyle e^{v}=y+c \)
Putting back \( v=\frac{x}{y} \):
\( \displaystyle e^{x/y}=y+c \)
Shortcut: Differentiate Option (2) with respect to y, treating x as a function of y:
\( \displaystyle e^{x/y}\cdot\frac{y\frac{dx}{dy}-x}{y^2}=1 \)
\( \displaystyle e^{x/y}\left(y\,dx-x\,dy\right)=y^2\,dy \)
This rearranges to \( ye^{x/y}dx=\left(xe^{x/y}+y^2\right)dy \), the given equation.
NCERT (2026-27 edition):
Differential Equations
