For differential equation \(y\,e^{\frac{x}{y}}\,dx = \left(x\,e^{\frac{x}{y}} + y^2\right)dy\), y(0) = 1, the value of x(e) is equal to:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → e
The question asks for x(e), so treat x as a function of y:
\( \displaystyle \frac{dx}{dy} = \frac{x\,e^{x/y} + y^2}{y\,e^{x/y}} \)
\( \displaystyle = \frac{x}{y} + y\,e^{-x/y} \)
The ratio x/y suggests the substitution x = vy, so
\( \displaystyle \frac{dx}{dy} = v + y\frac{dv}{dy} \)
Then
\( \displaystyle v + y\frac{dv}{dy} = v + y\,e^{-v} \)
\( \displaystyle \frac{dv}{dy} = e^{-v} \)
\(e^{v}\,dv = dy\)
Integrating both sides:
\(e^{v} = y + C\)
\(e^{x/y} = y + C\)
The condition y(0) = 1 means y = 1 when x = 0:
\(e^{0} = 1 + C\)
\(C = 0\)
So the particular solution is
\(e^{x/y} = y\)
\(x = y\log y\)
x(e) is the value of x when y = e:
\(x(e) = e\log e = e\)