$$ \sqrt{1-x^2}\differentialD y-\text{ydx}=0 $$

The given differential equation is:

\[
\sqrt{1-x^2} \, dy - y \, dx = 0
\]

This can be rewritten as:

\[
\sqrt{1 - x^2} \, dy = y \, dx
\]

To solve this differential equation, we can first rewrite it in a separable form:

\[
\frac{dy}{y} = \frac{dx}{\sqrt{1 - x^2}}
\]

Now, integrate both sides:

\[
\int \frac{1}{y} \, dy = \int \frac{1}{\sqrt{1 - x^2}} \, dx
\]

### Left-hand side:
\[
\int \frac{1}{y} \, dy = \ln |y|
\]

### Right-hand side:
The integral of $\frac{1}{\sqrt{1 - x^2}}$ is the inverse sine function, so:
\[
\int \frac{1}{\sqrt{1 - x^2}} \, dx = \arcsin(x)
\]

Now, combining both integrals:

\[
\ln |y| = \arcsin(x) + C
\]

Exponentiate both sides to solve for $y$:

\[
|y| = e^{\arcsin(x) + C} = e^C e^{\arcsin(x)}
\]

Let $A = e^C$, a constant. So, we have:

\[
y = A e^{\arcsin(x)}
\]

Thus, the general solution to the differential equation is:

\[
y = A e^{\arcsin(x)}
\]

---

Would you like more details on any step or have any further questions?

Here are five related questions for further exploration:

1. How do we find the particular solution if an initial condition is provided?
2. Can we use a different method to solve this differential equation, such as an exact method?
3. What is the domain of the solution $y = A e^{\arcsin(x)}$?
4. How does the inverse sine function behave near its critical points?
5. How would we solve if the equation were slightly different, such as $\sqrt{1 - x^2} dy + y dx = 0$?

**Tip:** Remember, inverse trigonometric functions often arise in differential equations involving square roots of quadratic expressions.

Solve the differential equation: $$ \sqrt{1-x^2}dy - ydx = 0 $$

Explore how to solve the differential equation $ \sqrt{1-x^2} dy - y dx = 0 $ using separation of variables and integration. This solution involves trigonometric functions and exponential expressions, providing a step-by-step walkthrough for undergraduate calculus students.

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