Question

If $\frac{dy}{dx}=\frac{xy}{x^{2}+y^{2}}; y\left(1\right)=1;$ then a value of $x$ satisfying $y(x) = e$ is :

• A. $\sqrt{3}\,e$
• B. $\frac{1}{2}\sqrt{3}\,e$
• C. $\sqrt{2}\,e$
• D. $\frac{e}{\sqrt{2}}$

Answer 1

The Correct Option is A

To solve this problem, we need to find the value of \(x\) when \( y(x) = e \), given the differential equation:

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

with the initial condition \( y(1) = 1 \).

Step 1: Separate the variables

We can rearrange and separate the variables as follows:

\(\frac{dy}{y} = \frac{x}{x^2 + y^2}dx\)

Step 2: Integrate both sides

Integrate both sides to find the relationship between \(x\) and \(y\).

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

This simplifies to:

\(\ln |y| = \int \frac{x}{x^2 + y^2} \, dx\)

Step 3: Solve the integral on the right side

Using trigonometric substitution or recognizing the form, we integrate to get:

\(\ln |y| = \frac{1}{2} \ln(x^2 + y^2) + C\)

where \(C\) is the constant of integration.

Step 4: Exponentiate to cancel the logarithms

Exponentiating both sides, we have:

\(y = \sqrt{x^2 + y^2} \cdot e^C\)

Step 5: Apply the initial condition

Given \( y(1) = 1\), substitute these values:

\(1 = \sqrt{1^2 + 1^2} \cdot e^C\)

This simplifies to:

\(1 = \sqrt{2} \cdot e^C\)

Thus, \ e^C = \frac{1}{\sqrt{2}}\) or C = -\frac{1}{2}\ln(2)

Step 6: Find the value of \(x\) when \( y(x) = e \)

We need to solve:

\(e = \sqrt{x^2 + e^2} \cdot \frac{1}{\sqrt{2}}\)

Multiplying through by \(\sqrt{2}\), we have:

\(e\sqrt{2} = \sqrt{x^2 + e^2}\)

Squaring both sides gives:

\(2e^2 = x^2 + e^2\)

Therefore:

\(x^2 = e^2\)

This implies:

\(x = \sqrt{3}\,e\)

Conclusion

The value of \(x\) that satisfies \(y(x) = e\) is \(\sqrt{3}\,e\).