Question

Solve \(y^{\prime} \sqrt{1-x^{2}}+\sqrt{1-y^{2}}=0\) explicitly.

Short answer

Question: Determine the explicit solution of the differential equation \(y^{\prime} \sqrt{1-x^{2}}+\sqrt{1-y^{2}}=0\). Answer: The explicit solution of the given differential equation is \(y(x) = \sin(\arcsin(x) + \arcsin(k))\), where \(k\) is a constant.

Step-by-step solution

Rewrite the given equation

To solve the given differential equation, let's first rewrite it in the form that isolates \(y^{\prime}\): $$ y^{\prime} \sqrt{1-x^{2}} = -\sqrt{1-y^{2}} $$ Now, we can divide both sides by \(\sqrt{1-x^{2}}\sqrt{1-y^{2}}\) to get: $$ y^{\prime} = -\frac{\sqrt{1-y^{2}}}{\sqrt{1-x^{2}}} $$

Separate the variables

To further simplify the equation and prepare it for integration, we will separate the variables. We can rewrite the equation as: $$ -\frac{dy}{\sqrt{1-y^{2}}} = \frac{dx}{\sqrt{1-x^{2}}} $$ Now, we have the two variables separated.

Integrate both sides

Let's integrate both sides of the equation with respect to their respective variables: $$ -\int\frac{dy}{\sqrt{1-y^{2}}} = \int\frac{dx}{\sqrt{1-x^{2}}} $$ The left side integral is a standard integral, which evaluates to \(-\arcsin(y) + C_1\). The right side integral is also a standard integral, which evaluates to \(\arcsin(x) + C_2\). Hence, we have: $$ -\arcsin(y) + C_1 = \arcsin(x) + C_2 $$

Solve for y(x)

Now, let's solve for y(x): $$ \arcsin(y) = \arcsin(x) + C_3, $$ where \(C_3 = C_2 - C_1\). To find y(x), we simply take the inverse sine of both sides: $$ y(x) = \sin(\arcsin(x) + C_3) = \sin(\arcsin(x) + \arcsin(k)) $$ where \(k = \sin(C_3)\) is a constant. This is the explicit solution to the given differential equation: $$ y(x) = \sin(\arcsin(x) + \arcsin(k)) $$