The solution of the differential equation y f | vedclass

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(B) Given the differential equation $y \frac{dy}{dx} + x = k$. Rearranging the terms,we get $y \frac{dy}{dx} = k - x$. Integrating both sides with res

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Circle

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(B) Given the differential equation $y \frac{dy}{dx} + x = k$. Rearranging the terms,we get $y \frac{dy}{dx} = k - x$. Integrating both sides with respect to $x$: $\int y \, dy = \int (k - x) \, dx$. This gives $\frac{y^2}{2} = kx - \frac{x^2}{2} + C$,where $C$ is the constant of integration. Multiplying by $2$,we get $y^2 = 2kx - x^2 + 2C$. Rearranging the terms: $x^2 - 2kx + y^2 = 2C$. Completing the square for $x$: $(x^2 - 2kx + k^2) + y^2 = 2C + k^2$. $(x - k)^2 + y^2 = 2C + k^2$. This is the equation of a circle in the form $(x - h)^2 + (y - k_0)^2 = r^2$,where the center is $(k, 0)$ and the radius is $\sqrt{2C + k^2}$. Therefore,the solution represents a circle.

The solution of the differential equation $y \frac{dy}{dx} + x = k$ represents . . . . . . .

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