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Question

(A) Given the differential equation: $\left(x+y \frac{dy}{dx}\right)=1$. Rearranging the terms,we get: $y \frac{dy}{dx} = 1 - x$. This can be written

Vedclass · Accepted Answer

$x^2+y^2=2x+C$

Vedclass · Answer

(A) Given the differential equation: $\left(x+y \frac{dy}{dx}\right)=1$. Rearranging the terms,we get: $y \frac{dy}{dx} = 1 - x$. This can be written as: $\frac{dy}{dx} = \frac{1-x}{y}$. Separating the variables: $y \, dy = (1-x) \, dx$. Integrating both sides: $\int y \, dy = \int (1-x) \, dx$. $\frac{y^2}{2} = x - \frac{x^2}{2} + C_1$. Multiplying by $2$: $y^2 = 2x - x^2 + 2C_1$. Rearranging: $x^2 + y^2 - 2x = C$ (where $C = 2C_1$). Thus,the general solution is $x^2 + y^2 - 2x = C$.

Find the general solution of the differential equation: $\left(x+y \frac{dy}{dx}\right)=1$.

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