The solution of the equation frac{dy}{dx} = f | vedclass

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(B) Given the homogeneous differential equation $\frac{dy}{dx} = \frac{x + y}{x - y}$.Substitute $y = vx$,which implies $\frac{dy}{dx} = v + x\frac{dv

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$c(x^2 + y^2)^{1/2} = e^{	an^{-1}(y/x)}$

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(B) Given the homogeneous differential equation $\frac{dy}{dx} = \frac{x + y}{x - y}$.Substitute $y = vx$,which implies $\frac{dy}{dx} = v + x\frac{dv}{dx}$.Substituting these into the equation: $v + x\frac{dv}{dx} = \frac{x + vx}{x - vx} = \frac{1 + v}{1 - v}$.Rearranging terms: $x\frac{dv}{dx} = \frac{1 + v}{1 - v} - v = \frac{1 + v - v + v^2}{1 - v} = \frac{1 + v^2}{1 - v}$.Separating variables: $\frac{dx}{x} = \frac{1 - v}{1 + v^2} dv = \left( \frac{1}{1 + v^2} - \frac{v}{1 + v^2} ight) dv$.Integrating both sides: $\int \frac{dx}{x} = \int \frac{1}{1 + v^2} dv - \int \frac{v}{1 + v^2} dv$.$\ln|x| = 	an^{-1}(v) - \frac{1}{2} \ln(1 + v^2) + C$.Substitute $v = \frac{y}{x}$: $\ln|x| = 	an^{-1}(\frac{y}{x}) - \frac{1}{2} \ln(1 + \frac{y^2}{x^2}) + \ln|c|$.$\ln|x| = 	an^{-1}(\frac{y}{x}) - \frac{1}{2} \ln(\frac{x^2 + y^2}{x^2}) + \ln|c|$.$\ln|x| = 	an^{-1}(\frac{y}{x}) - \frac{1}{2} [\ln(x^2 + y^2) - \ln(x^2)] + \ln|c|$.$\ln|x| = 	an^{-1}(\frac{y}{x}) - \frac{1}{2} \ln(x^2 + y^2) + \ln|x| + \ln|c|$.$0 = 	an^{-1}(\frac{y}{x}) - \ln((x^2 + y^2)^{1/2}) + \ln|c|$.$\ln((x^2 + y^2)^{1/2}) = 	an^{-1}(\frac{y}{x}) + \ln|c|$.Taking the exponential of both sides: $(x^2 + y^2)^{1/2} = c e^{	an^{-1}(y/x)}$ or $c(x^2 + y^2)^{1/2} = e^{	an^{-1}(y/x)}$.

The solution of the equation $\frac{dy}{dx} = \frac{x + y}{x - y}$ is

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