1. Given differential equation

$\frac{dy}{dx} = \frac{xy}{x^2+y^2}, \qquad y(1)=1$

We need to find $x$ such that $y(x)=e$.

2. Invert the differential equation

It is convenient to write $x$ as a function of $y$:

$\frac{dx}{dy} = \frac{x^2+y^2}{xy} = \frac{x}{y}+\frac{y}{x}$

Now let

$u = x^2$

Then

$\frac{du}{dy} = 2x\frac{dx}{dy}$

Using the expression for $\frac{dx}{dy}$,

$\frac{du}{dy} = 2x\left(\frac{x}{y}+\frac{y}{x}\right)=\frac{2x^2}{y}+2y$

Since $u=x^2$,

$\frac{du}{dy} = \frac{2u}{y}+2y$

So we get the linear differential equation

$\frac{du}{dy}-\frac{2}{y}u = 2y$

3. Solve the linear equation

The integrating factor is

$\text{I.F.}=e^{\int -2/y\,dy}=e^{-2\ln y}=y^{-2}$

Multiplying throughout by $y^{-2}$:

$y^{-2}\frac{du}{dy}-2y^{-3}u = 2y\cdot y^{-2}=\frac{2}{y}$

The left side becomes:

$\frac{d}{dy}(u y^{-2}) = \frac{2}{y}$

Integrate:

$u y^{-2} = 2\ln y + C$

Hence

$u = y^2(2\ln y + C)$

Since $u=x^2$,

$x^2 = y^2(2\ln y + C)$

4. Use the initial condition

Given $y(1)=1$, so when $x=1$, $y=1$:

$1^2 = 1^2(2\ln 1 + C)$

Since $\ln 1=0$,

$1=C$

Therefore,

$x^2 = y^2(2\ln y + 1)$

5. Substitute $y=e$

For $y=e$,

$x^2 = e^2(2\ln e + 1)=e^2(2\cdot 1 +1)=3e^2$

So

$x = \pm \sqrt{3}\,e$

From the options, the listed value is the positive one:

$x=\sqrt{3}\,e$

6. Check options

• A: $\sqrt{2}e$ ❌
• B: $\frac{\sqrt{3}}{2}e$ ❌
• C: $\frac{e}{\sqrt{2}}$ ❌
• D: $\sqrt{3}e$ ✅

So the correct option is D.

7. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

They agree.