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

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

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

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

$$a)\frac{dy}{1+y^2}+\frac{e^{x}dx}{1+{\left(e^x\right)}^2}=0$$$$\int \frac{dy}{1+y^2}+\int \frac{e^{x}dx}{1+{\left(e^x\right)}^2}=c$$$${tan}^{-1}\left(y\right)+{tan}^{-1}\left(e^x\right)=c$$$${tan}^{-1}\left(\frac{y+e^x}{1-y.e^x}\right)=c$$$$\frac{y+e^x}{1-{ye}^x}=tanc=C_1$$$$\frac{y+e^x=C_1(1-{ye}^x)}{}$$

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