Question-110227

Question Number 110227 by mohammad17 last updated on 27/Aug/20

Commented by mohammad17 last updated on 27/Aug/20

$${help}\:{me}\:{sir} \\ $$

Answered by bobhans last updated on 28/Aug/20

(a) we obtain y= ±(√((1−x^3 )/(3x))) . let ∅(x)=(√((1−x^3 )/(3x))) to see if it is an implicit solution . Since (d∅/dx) = (((−6x^3 −3)(√(3x)))/(18x^2 (√(1−x^3 )))) (d∅/dx) = −(((2x^3 +1)(√(3x)))/(6x^2 (√(1−x^3 )))) are defined on (0,1) substituting them into diff eq yields 2xy(−(((2x^3 +1)(√(3x)))/(6x^2 (√(1−x^3 )))))+x^2 +y^2 = 2x((√((1−x^3 )/(3x))))(−(((2x^3 +1)(√(3x)))/(6x^2 (√(1−x^3 )))))+x^2 +((1−x^3 )/(3x))= −(((2x^3 +1))/(3x))+((3x^3 +1−x^3 )/(3x)) = −(((2x^3 +1)/(3x)))+(((2x^3 +1)/(3x)))=0 so x^3 +3xy^2 =1 is an implicit solution

$$\left({a}\right)\:{we}\:{obtain}\:{y}=\:\pm\sqrt{\frac{\mathrm{1}−{x}^{\mathrm{3}} }{\mathrm{3}{x}}}\:. \\ $$$${let}\:\emptyset\left({x}\right)=\sqrt{\frac{\mathrm{1}−{x}^{\mathrm{3}} }{\mathrm{3}{x}}}\:{to}\:{see}\:{if}\:{it}\:{is}\:{an}\:{implicit} \\ $$$${solution}\:.\:{Since}\:\frac{{d}\emptyset}{{dx}}\:=\:\frac{\left(−\mathrm{6}{x}^{\mathrm{3}} −\mathrm{3}\right)\sqrt{\mathrm{3}{x}}}{\mathrm{18}{x}^{\mathrm{2}} \:\sqrt{\mathrm{1}−{x}^{\mathrm{3}} }} \\ $$$$\frac{{d}\emptyset}{{dx}}\:=\:−\frac{\left(\mathrm{2}{x}^{\mathrm{3}} +\mathrm{1}\right)\sqrt{\mathrm{3}{x}}}{\mathrm{6}{x}^{\mathrm{2}} \sqrt{\mathrm{1}−{x}^{\mathrm{3}} }}\:{are}\:{defined}\:{on}\:\left(\mathrm{0},\mathrm{1}\right) \\ $$$${substituting}\:{them}\:{into}\:{diff}\:{eq}\:{yields} \\ $$$$\mathrm{2}{xy}\left(−\frac{\left(\mathrm{2}{x}^{\mathrm{3}} +\mathrm{1}\right)\sqrt{\mathrm{3}{x}}}{\mathrm{6}{x}^{\mathrm{2}} \sqrt{\mathrm{1}−{x}^{\mathrm{3}} }}\right)+{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \:= \\ $$$$\mathrm{2}{x}\left(\sqrt{\frac{\mathrm{1}−{x}^{\mathrm{3}} }{\mathrm{3}{x}}}\right)\left(−\frac{\left(\mathrm{2}{x}^{\mathrm{3}} +\mathrm{1}\right)\sqrt{\mathrm{3}{x}}}{\mathrm{6}{x}^{\mathrm{2}} \sqrt{\mathrm{1}−{x}^{\mathrm{3}} }}\right)+{x}^{\mathrm{2}} +\frac{\mathrm{1}−{x}^{\mathrm{3}} }{\mathrm{3}{x}}= \\ $$$$−\frac{\left(\mathrm{2}{x}^{\mathrm{3}} +\mathrm{1}\right)}{\mathrm{3}{x}}+\frac{\mathrm{3}{x}^{\mathrm{3}} +\mathrm{1}−{x}^{\mathrm{3}} }{\mathrm{3}{x}}\:=\:−\left(\frac{\mathrm{2}{x}^{\mathrm{3}} +\mathrm{1}}{\mathrm{3}{x}}\right)+\left(\frac{\mathrm{2}{x}^{\mathrm{3}} +\mathrm{1}}{\mathrm{3}{x}}\right)=\mathrm{0} \\ $$$${so}\:{x}^{\mathrm{3}} +\mathrm{3}{xy}^{\mathrm{2}} =\mathrm{1}\:{is}\:{an}\:{implicit}\:{solution} \\ $$$$ \\ $$

Commented by mohammad17 last updated on 28/Aug/20

$${thank}\:{you}\:{sir}\:{can}\:{you}\:{help}\:{me}\:{in}\:\left({b}\right) \\ $$

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