Question-46358

Question Number 46358 by peter frank last updated on 24/Oct/18

Answered by tanmay.chaudhury50@gmail.com last updated on 24/Oct/18

b)(x+y)^2 (xdy+ydx)=xy(dx+dy) (x+y)^2 d(xy)=xy(dx+dy) ((d(xy))/(xy))=((d(x+y))/((x+y)^2 )) ∫((d(xy))/(xy))=∫((d(x+y))/((x+y)^2 )) ln(xy)=((−1)/((x+y)))+lnc ln(((xy)/c))=((−1)/((x+y))) xy=ce^((−1)/((x+y)))

$$\left.{b}\right)\left({x}+{y}\right)^{\mathrm{2}} \left({xdy}+{ydx}\right)={xy}\left({dx}+{dy}\right) \\ $$$$\left({x}+{y}\right)^{\mathrm{2}} {d}\left({xy}\right)={xy}\left({dx}+{dy}\right) \\ $$$$\frac{{d}\left({xy}\right)}{{xy}}=\frac{{d}\left({x}+{y}\right)}{\left({x}+{y}\right)^{\mathrm{2}} } \\ $$$$\int\frac{{d}\left({xy}\right)}{{xy}}=\int\frac{{d}\left({x}+{y}\right)}{\left({x}+{y}\right)^{\mathrm{2}} } \\ $$$${ln}\left({xy}\right)=\frac{−\mathrm{1}}{\left({x}+{y}\right)}+{lnc} \\ $$$${ln}\left(\frac{{xy}}{{c}}\right)=\frac{−\mathrm{1}}{\left({x}+{y}\right)} \\ $$$${xy}={ce}^{\frac{−\mathrm{1}}{\left({x}+{y}\right)}} \\ $$$$ \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 24/Oct/18

a)(dy/(1+y^2 ))+((e^x dx)/(1+(e^x )^2 ))=0 ∫(dy/(1+y^2 ))+∫((e^x dx)/(1+(e^x )^2 ))=c tan^(−1) (y)+tan^(−1) (e^x )=c tan^(−1) (((y+e^x )/(1−y.e^x )))=c ((y+e^x )/(1−ye^x ))=tanc=C_1 ((y+e^x =C_1 (1−ye^x ))/)

$$\left.{a}\right)\frac{{dy}}{\mathrm{1}+{y}^{\mathrm{2}} }+\frac{{e}^{{x}} {dx}}{\mathrm{1}+\left({e}^{{x}} \right)^{\mathrm{2}} }=\mathrm{0} \\ $$$$\int\frac{{dy}}{\mathrm{1}+{y}^{\mathrm{2}} }+\int\frac{{e}^{{x}} {dx}}{\mathrm{1}+\left({e}^{{x}} \right)^{\mathrm{2}} }={c} \\ $$$${tan}^{−\mathrm{1}} \left({y}\right)+{tan}^{−\mathrm{1}} \left({e}^{{x}} \right)={c} \\ $$$${tan}^{−\mathrm{1}} \left(\frac{{y}+{e}^{{x}} }{\mathrm{1}−{y}.{e}^{{x}} }\right)={c} \\ $$$$\frac{{y}+{e}^{{x}} }{\mathrm{1}−{ye}^{{x}} }={tanc}={C}_{\mathrm{1}} \\ $$$$\frac{{y}+{e}^{{x}} ={C}_{\mathrm{1}} \left(\mathrm{1}−{ye}^{{x}} \right)}{} \\ $$

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