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Solve the differential equation. \[ \frac{d y}{d x}=x \sqrt{y} \] for \( y \neq 0 \)

Ask by Simmons Long. in the United States
Mar 16,2025

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Answer

The solution to the differential equation \( \frac{dy}{dx} = x \sqrt{y} \) is: \[ y = \left(\frac{x^2}{4} + C\right)^2, \] where \( C \) is an arbitrary constant.

Solution

We start with the differential equation \[ \frac{dy}{dx} = x \sqrt{y}. \] Since \( y \neq 0 \), we can separate the variables. Dividing both sides by \(\sqrt{y}\), we have \[ \frac{1}{\sqrt{y}}\,dy = x\,dx. \] Now, we integrate both sides: \[ \int \frac{1}{\sqrt{y}}\,dy = \int x\,dx. \] The left-hand side integrates as follows: \[ \int y^{-\frac{1}{2}}\,dy = 2\sqrt{y} + C_1, \] and the right-hand side is \[ \int x\,dx = \frac{x^2}{2} + C_2. \] We can combine the constants \( C_1 \) and \( C_2 \) into a single constant \( C \) (since both are arbitrary). Thus, we write: \[ 2\sqrt{y} = \frac{x^2}{2} + C. \] To solve for \(\sqrt{y}\), divide both sides by 2: \[ \sqrt{y} = \frac{x^2}{4} + C_0, \] where \( C_0 = \frac{C}{2} \) is an arbitrary constant. Finally, to express \( y \) explicitly, square both sides: \[ y = \left(\frac{x^2}{4} + C_0\right)^2. \] Thus, the general solution to the differential equation is \[ \boxed{y = \left(\frac{x^2}{4} + C_0\right)^2, \quad \text{where } C_0 \text{ is an arbitrary constant.}} \]

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To solve the differential equation \(\frac{d y}{d x} = x \sqrt{y}\), we can separate the variables. Rearranging gives us: \[ \frac{1}{\sqrt{y}} \, dy = x \, dx. \] Next, integrate both sides. The left side integrates to \(2\sqrt{y}\) and the right side integrates to \(\frac{1}{2}x^2\). So we have: \[ 2\sqrt{y} = \frac{1}{2} x^2 + C, \] where \(C\) is a constant of integration. Divide through by 2 to simplify: \[ \sqrt{y} = \frac{1}{4} x^2 + \frac{C}{2}. \] Now, squaring both sides gives us: \[ y = \left(\frac{1}{4} x^2 + \frac{C}{2}\right)^2. \] This represents the general solution to the differential equation.

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