The given differential equation is:

\[
\frac{dy}{dx} = \frac{y}{x} + x \sin\left(\frac{y}{x}\right)
\]

### Solution:

To solve this equation, let's attempt a substitution to simplify the equation. Define:

\[
v = \frac{y}{x}
\]

This implies:

\[
y = vx \quad \text{and} \quad \frac{dy}{dx} = v + x\frac{dv}{dx}
\]

Substituting these into the original equation:

\[
v + x\frac{dv}{dx} = v + x \sin(v)
\]

Canceling the \(v\) terms from both sides:

\[
x \frac{dv}{dx} = x \sin(v)
\]

Dividing by \(x\) on both sides:

\[
\frac{dv}{dx} = \sin(v)
\]

Now, separate variables:

\[
\frac{dv}{\sin(v)} = dx
\]

Integrating both sides:

\[
\int \frac{dv}{\sin(v)} = \int dx
\]

The integral of \(\frac{1}{\sin(v)}\) is \(\ln|\tan(v/2)|\), so we get:

\[
\ln|\tan(v/2)| = x + C
\]

Exponentiating both sides to remove the logarithm:

\[
|\tan(v/2)| = e^{x+C} = A e^x \quad \text{where} \quad A = e^C
\]

Thus:

\[
\tan(v/2) = \pm A e^x
\]

Finally, recall that \(v = \frac{y}{x}\), so the solution is:

\[
\tan\left(\frac{y}{2x}\right) = A e^x
\]

This is the implicit solution to the given differential equation.

---

Would you like further details or have any questions on this solution?

### Related Questions:
1. How can you verify if this solution satisfies the original equation?
2. What is the significance of choosing the positive or negative sign in the solution?
3. Can the solution be expressed explicitly in terms of \(y\)?
4. How does the constant \(A\) affect the behavior of the solution?
5. Can this method be applied to other types of differential equations?

### Tip:
When facing a differential equation that involves \(\frac{y}{x}\), substituting \(v = \frac{y}{x}\) is often a useful technique to simplify the problem.

This page provides a detailed step-by-step solution to the differential equation dy/dx = y/x + x sin(y/x) using substitution methods. It explores core concepts in differential equations, including substitution and separation of variables, and explains how to derive the implicit solution. Suitable for undergraduate students studying calculus or differential equations.

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