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e-x-y-x-2-y-2-find-the-expression-for-dy-dx-




Question Number 150839 by mathdanisur last updated on 15/Aug/21
e^x  + y = x^2 y^2   find the expression for  (dy/dx)
$$\mathrm{e}^{\boldsymbol{\mathrm{x}}} \:+\:\mathrm{y}\:=\:\mathrm{x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{expression}\:\mathrm{for}\:\:\frac{\mathrm{dy}}{\mathrm{dx}} \\ $$
Answered by peter frank last updated on 15/Aug/21
e^x +(dy/dx)=x^2 2y(dy/dx)+y^2 2x  e^x −2xy^2 =(dy/dx)(2x^2 y−1)  (dy/dx)=((e^x −2xy^2 )/(2x^2 y−1))
$${e}^{{x}} +\frac{{dy}}{{dx}}={x}^{\mathrm{2}} \mathrm{2}{y}\frac{{dy}}{{dx}}+{y}^{\mathrm{2}} \mathrm{2}{x} \\ $$$${e}^{{x}} −\mathrm{2}{xy}^{\mathrm{2}} =\frac{{dy}}{{dx}}\left(\mathrm{2}{x}^{\mathrm{2}} {y}−\mathrm{1}\right) \\ $$$$\frac{{dy}}{{dx}}=\frac{{e}^{{x}} −\mathrm{2}{xy}^{\mathrm{2}} }{\mathrm{2}{x}^{\mathrm{2}} {y}−\mathrm{1}} \\ $$
Commented by mathdanisur last updated on 15/Aug/21
Thank you ser
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{ser} \\ $$
Answered by amin96 last updated on 15/Aug/21
((d(e^x ))/dx)+(dy/dx)=x^2 ((d(y^2 ))/dx)+y^2 ((d(x^2 ))/dx)  e^x +(dy/dx)=2x^2 y(dy/dx)+2xy^2   ⇒  { y=f(x)  ((d(y^n ))/dx)=ny^(n−1) (dy/dx)}  e^x +(dy/dx)(1−2x^2 y)=2xy^2   (dy/dx)=((2xy^2 −e^x )/(1−2x^2 y))
$$\frac{{d}\left({e}^{{x}} \right)}{{dx}}+\frac{{dy}}{{dx}}={x}^{\mathrm{2}} \frac{{d}\left({y}^{\mathrm{2}} \right)}{{dx}}+{y}^{\mathrm{2}} \frac{{d}\left({x}^{\mathrm{2}} \right)}{{dx}} \\ $$$${e}^{{x}} +\frac{{dy}}{{dx}}=\mathrm{2}{x}^{\mathrm{2}} {y}\frac{{dy}}{{dx}}+\mathrm{2}{xy}^{\mathrm{2}} \:\:\Rightarrow\:\:\left\{\:{y}={f}\left({x}\right)\:\:\frac{{d}\left({y}^{{n}} \right)}{{dx}}={ny}^{{n}−\mathrm{1}} \frac{{dy}}{{dx}}\right\} \\ $$$${e}^{{x}} +\frac{{dy}}{{dx}}\left(\mathrm{1}−\mathrm{2}{x}^{\mathrm{2}} {y}\right)=\mathrm{2}{xy}^{\mathrm{2}} \\ $$$$\frac{{dy}}{{dx}}=\frac{\mathrm{2}{xy}^{\mathrm{2}} −{e}^{{x}} }{\mathrm{1}−\mathrm{2}{x}^{\mathrm{2}} {y}} \\ $$
Commented by mathdanisur last updated on 15/Aug/21
Thank you Ser
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Ser} \\ $$

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