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Solve: e^xsqrt(1-y^2)dx+y/x dy=0...

Solve: `e^xsqrt(1-y^2)dx+y/x dy=0`

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To solve the differential equation \( e^x \sqrt{1 - y^2} \, dx + \frac{y}{x} \, dy = 0 \), we can follow these steps: ### Step 1: Rearranging the Equation We start by rearranging the equation to separate the variables \(x\) and \(y\): \[ e^x \sqrt{1 - y^2} \, dx = -\frac{y}{x} \, dy \] ### Step 2: Separating Variables Next, we can rewrite the equation in a form that separates the variables: \[ \frac{e^x \sqrt{1 - y^2}}{y} \, dx = -\frac{1}{x} \, dy \] Now, we can multiply both sides by \(x\): \[ x e^x \sqrt{1 - y^2} \, dx = -y \, dy \] ### Step 3: Integrating Both Sides Now we can integrate both sides. We will integrate the left side with respect to \(x\) and the right side with respect to \(y\): \[ \int x e^x \, dx = -\int y \, dy \] ### Step 4: Solving the Left Integral To solve the left integral, we can use integration by parts. Let: - \(u = x\) and \(dv = e^x \, dx\) - Then, \(du = dx\) and \(v = e^x\) Using integration by parts: \[ \int u \, dv = uv - \int v \, du \] We have: \[ \int x e^x \, dx = x e^x - \int e^x \, dx = x e^x - e^x + C_1 \] ### Step 5: Solving the Right Integral Now, we integrate the right side: \[ -\int y \, dy = -\frac{y^2}{2} + C_2 \] ### Step 6: Equating Both Integrals Now we equate both sides: \[ x e^x - e^x = -\frac{y^2}{2} + C \] where \(C = C_2 - C_1\). ### Step 7: Rearranging the Equation Rearranging gives us: \[ x e^x - e^x + \frac{y^2}{2} = C \] ### Step 8: Final Form This is the implicit solution of the differential equation. We can express it as: \[ x e^x - e^x + \frac{y^2}{2} = C \] ### Summary The solution to the differential equation \( e^x \sqrt{1 - y^2} \, dx + \frac{y}{x} \, dy = 0 \) is given by: \[ x e^x - e^x + \frac{y^2}{2} = C \]

To solve the differential equation \( e^x \sqrt{1 - y^2} \, dx + \frac{y}{x} \, dy = 0 \), we can follow these steps: ### Step 1: Rearranging the Equation We start by rearranging the equation to separate the variables \(x\) and \(y\): \[ e^x \sqrt{1 - y^2} \, dx = -\frac{y}{x} \, dy \] ...
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