Solve the differential equation frac{dy}{dx} | vedclass

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Question

(C) Given the differential equation: $\frac{dy}{dx} = \frac{y(1+x)}{-x(1+y)}$.Separate the variables $x$ and $y$:$\frac{1+y}{y} dy = -\frac{1+x}{x} dx

Vedclass · Accepted Answer

$x+y+\log(xy)=c$

Vedclass · Answer

(C) Given the differential equation: $\frac{dy}{dx} = \frac{y(1+x)}{-x(1+y)}$.Separate the variables $x$ and $y$:$\frac{1+y}{y} dy = -\frac{1+x}{x} dx$.This can be written as:$(\frac{1}{y} + 1) dy = -(\frac{1}{x} + 1) dx$.Integrate both sides:$\int (\frac{1}{y} + 1) dy = -\int (\frac{1}{x} + 1) dx$.$\log|y| + y = -(\log|x| + x) + C$.$\log|y| + y = -\log|x| - x + C$.Rearranging the terms:$x + y + \log|x| + \log|y| = C$.Using the property $\log a + \log b = \log(ab)$:$x + y + \log|xy| = C$.Thus,the correct option is $C$.

Solve the differential equation $\frac{dy}{dx} = \frac{y(1+x)}{-x(1+y)}$.

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