Question Number 25520 by Mahesh Andiboina last updated on 11/Dec/17
Commented by Mahesh Andiboina last updated on 11/Dec/17
$$\mathrm{plzz}\:\mathrm{ans}\:\mathrm{both} \\ $$
Answered by prakash jain last updated on 11/Dec/17
$$\mathrm{15} \\ $$$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} −\mathrm{3}{axy}=\mathrm{0} \\ $$$$\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{y}^{\mathrm{2}} \frac{{dy}}{{dx}}−\mathrm{3}{a}\left({y}+{x}\frac{{dy}}{{dx}}\right)=\mathrm{0} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} \frac{{dy}}{{dx}}−{ay}−{ax}\frac{{dy}}{{dx}}=\mathrm{0} \\ $$$$\left({y}^{\mathrm{2}} −{ax}\right)\frac{{dy}}{{dx}}={ay}−{x}^{\mathrm{2}} \\ $$$$\frac{{dy}}{{dx}}=\frac{{ay}−{x}^{\mathrm{2}} }{{y}^{\mathrm{2}} −{ax}} \\ $$
Answered by prakash jain last updated on 11/Dec/17
$$\mathrm{14} \\ $$$$\mathrm{log}\:\left({x}/{y}\right)={x}+{y} \\ $$$$\mathrm{log}\:{x}−\mathrm{log}\:{y}={x}+{y} \\ $$$$\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{{y}}\frac{{dy}}{{dx}}=\mathrm{1}+\frac{{dy}}{{dx}} \\ $$$$\frac{\mathrm{1}}{{x}}−\mathrm{1}=\left(\mathrm{1}+\frac{\mathrm{1}}{{y}}\right)\frac{{dy}}{{dx}} \\ $$$$\frac{\mathrm{1}−{x}}{{x}}=\left(\frac{\mathrm{1}+{y}}{{y}}\right)\frac{{dy}}{{dx}} \\ $$$$\frac{{dy}}{{dx}}=\frac{{y}\left(\mathrm{1}−{x}\right)}{{x}\left(\mathrm{1}+{y}\right)} \\ $$