If x^alpha frac{dy}{dx} = y^beta(gamma log x | vedclass

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(A) The given differential equation is $x^\alpha \frac{dy}{dx} = y^\beta(\gamma \log x + \delta \log y + 1)$.For the equation to be homogeneous,the de

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$\alpha = \beta$ and $\gamma = -\delta$

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(A) The given differential equation is $x^\alpha \frac{dy}{dx} = y^\beta(\gamma \log x + \delta \log y + 1)$.For the equation to be homogeneous,the degree of the numerator and denominator must be the same,or the function must be expressible as a function of $\frac{y}{x}$.Rewriting the equation: $\frac{dy}{dx} = \frac{y^\beta}{x^\alpha} (\gamma \log x + \delta \log y + 1)$.For homogeneity,the powers of $x$ and $y$ must balance such that the expression depends only on the ratio $\frac{y}{x}$.This requires $\alpha = \beta$.Substituting $\alpha = \beta$,we get $\frac{dy}{dx} = (\frac{y}{x})^\alpha (\gamma \log x + \delta \log y + 1) = (\frac{y}{x})^\alpha (\gamma \log x + \delta \log y + \delta \log x - \delta \log x + 1)$.For the term to be a function of $\frac{y}{x}$,the $\log x$ terms must cancel out,which happens when $\gamma + \delta = 0$,i.e.,$\gamma = -\delta$.

If $x^\alpha \frac{dy}{dx} = y^\beta(\gamma \log x + \delta \log y + 1)$ is a homogeneous differential equation,then

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