Question Number 192545 by peter frank last updated on 20/May/23
Answered by BaliramKumar last updated on 20/May/23
$$\mathrm{plnx}+\mathrm{qlny}=\:\left(\mathrm{p}+\mathrm{q}\right)\mathrm{ln}\left(\mathrm{x}+\mathrm{y}\right) \\ $$$$\frac{\mathrm{p}}{\mathrm{x}}−\frac{\mathrm{p}+\mathrm{q}}{\mathrm{x}+\mathrm{y}}=\:\left\{\frac{\mathrm{p}+\mathrm{q}}{\mathrm{x}+\mathrm{y}}−\frac{\mathrm{q}}{\mathrm{y}}\right\}\frac{\mathrm{dy}}{\mathrm{dx}} \\ $$$$\frac{\mathrm{py}}{\mathrm{x}}\frac{−\mathrm{qx}}{}=\:\left\{\frac{\mathrm{py}−}{}\frac{\mathrm{qx}}{\mathrm{y}}\right\}\frac{\mathrm{dy}}{\mathrm{dx}} \\ $$$$\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{y}}{\mathrm{x}} \\ $$$$ \\ $$
Answered by aleks041103 last updated on 21/May/23
$${x}^{{p}} {y}^{{q}} =\left({x}+{y}\right)^{{p}+{q}} =\left(\mathrm{1}+\frac{{y}}{{x}}\right)^{{p}+{q}} {x}^{{p}+{q}} \\ $$$$\Rightarrow\left(\frac{{y}}{{x}}\right)^{{p}} =\left(\mathrm{1}+\frac{{y}}{{x}}\right)^{{p}+{q}} \\ $$$$\Rightarrow{k}=\left(\mathrm{1}+{k}\right)^{\mathrm{1}+\frac{{q}}{{p}}} \:\Rightarrow\:{k}={const}=\frac{{y}}{{x}}\Rightarrow{y}={kx} \\ $$$$\Rightarrow\frac{{dy}}{{dx}}={k}=\frac{{y}}{{x}} \\ $$
Answered by Spillover last updated on 21/May/23
$${x}^{{p}} {y}^{{q}} =\left({x}+{y}\right)^{{p}+{q}} \: \\ $$$$\mathrm{ln}\:\left({x}^{{p}} {y}^{{q}} \right)=\mathrm{ln}\:\left({x}+{y}\right)^{{p}+{q}} \: \\ $$$${p}\mathrm{ln}\:{x}+{q}\mathrm{ln}\:{y}=\left({p}+{q}\right)\mathrm{ln}\:\left({x}+{y}\right) \\ $$$$\frac{{p}}{{x}}+\frac{{q}}{{y}}\frac{{dy}}{{dx}}=\frac{\left({p}+{q}\right)\left(\mathrm{1}+\frac{{dy}}{{dx}}\right)}{{x}+{y}} \\ $$$$\frac{{py}+{qx}\frac{{dy}}{{dx}}}{{xy}}=\frac{\left({p}+{q}\right)\left(\mathrm{1}+\frac{{dy}}{{dx}}\right)}{{x}+{y}} \\ $$$$\frac{{dy}}{{dx}}=\frac{\left({p}+{q}\right){xy}−\left({x}+{y}\right){py}}{\left({x}+{y}\right){qx}−\left({p}+{q}\right){xy}} \\ $$$$\frac{{dy}}{{dx}}=\frac{−{py}^{\mathrm{2}} +{qxy}}{−{pxy}+{qx}^{\mathrm{2}} } \\ $$$$\frac{{dy}}{{dx}}=\frac{{y}}{{x}} \\ $$