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Question Number 82521 by jagoll last updated on 22/Feb/20

given (dy/dx) = xy^3 Determine the concavity of all solution curves for the given differential equation in quadrant iv.

$${given}\:\frac{{dy}}{{dx}}\:=\:{xy}^{\mathrm{3}} \\ $$$${Determine}\:{the}\:{concavity}\:{of}\:{all} \\ $$$${solution}\:{curves}\:{for}\:{the}\:{given}\: \\ $$$${differential}\:{equation}\:{in}\: \\ $$$${quadrant}\:{iv}.\: \\ $$

Commented by jagoll last updated on 22/Feb/20

someone maybe help me

$${someone}\:{maybe}\:{help}\:{me} \\ $$

Commented by mr W last updated on 22/Feb/20

(dy/dx)=xy^3 (d^2 y/dx^2 )=y^3 +3xy^2 (dy/dx)=y^3 (1+3x^2 y^2 ) in quadrant iii and iv: y<0 ⇒ y^3 (1+3x^2 y^2 )<0 ⇒ (d^2 y/dx^2 )<0 ⇒all solution curves are concave here! we even don′t need to solve the d. eqn.

$$\frac{{dy}}{{dx}}={xy}^{\mathrm{3}} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }={y}^{\mathrm{3}} +\mathrm{3}{xy}^{\mathrm{2}} \frac{{dy}}{{dx}}={y}^{\mathrm{3}} \left(\mathrm{1}+\mathrm{3}{x}^{\mathrm{2}} {y}^{\mathrm{2}} \right) \\ $$$${in}\:{quadrant}\:{iii}\:{and}\:{iv}:\:\:{y}<\mathrm{0} \\ $$$$\Rightarrow\:{y}^{\mathrm{3}} \left(\mathrm{1}+\mathrm{3}{x}^{\mathrm{2}} {y}^{\mathrm{2}} \right)<\mathrm{0} \\ $$$$\Rightarrow\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }<\mathrm{0} \\ $$$$\Rightarrow{all}\:{solution}\:{curves}\:{are}\:{concave}\:{here}! \\ $$$$ \\ $$$${we}\:{even}\:{don}'{t}\:{need}\:{to}\:{solve}\:{the}\:{d}.\:{eqn}. \\ $$

Commented by mr W last updated on 22/Feb/20

Commented by mr W last updated on 22/Feb/20

these are some examples from the solution curves.

$${these}\:{are}\:{some}\:{examples}\:{from}\:{the} \\ $$$${solution}\:{curves}. \\ $$

Commented by jagoll last updated on 22/Feb/20

(d^2 y/dx^2 ) = y^3 + 3xy^2 (dy/(dx )) sir?

$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:=\:{y}^{\mathrm{3}} \:+\:\mathrm{3}{xy}^{\mathrm{2}} \:\frac{{dy}}{{dx}\:} \\ $$$${sir}? \\ $$

Commented by mr W last updated on 22/Feb/20

yes sir! now it′s fixed. thanks alot!

$${yes}\:{sir}!\:{now}\:{it}'{s}\:{fixed}.\:{thanks}\:{alot}! \\ $$

Commented by john santu last updated on 22/Feb/20

∫ (dy/y^3 ) = ∫ xdx ⇒ −(1/(2y^2 )) = (1/2)x^2 −(1/(2 ))C y^2 = (1/(C−x^2 )) ⇒ y = ± (1/(√(C−x^2 ))) (d^2 y/(dx^2 )) = y^3 (1+3x^2 y^2 ) < 0

$$\int\:\frac{{dy}}{{y}^{\mathrm{3}} }\:=\:\int\:{xdx}\:\Rightarrow\:−\frac{\mathrm{1}}{\mathrm{2}{y}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}\:}{C} \\ $$$${y}^{\mathrm{2}} \:=\:\frac{\mathrm{1}}{{C}−{x}^{\mathrm{2}} \:}\:\Rightarrow\:{y}\:=\:\pm\:\frac{\mathrm{1}}{\sqrt{{C}−{x}^{\mathrm{2}} }}\: \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} \:}\:=\:{y}^{\mathrm{3}} \:\left(\mathrm{1}+\mathrm{3}{x}^{\mathrm{2}} {y}^{\mathrm{2}} \right)\:<\:\mathrm{0}\: \\ $$$$ \\ $$

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