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Question Number 169389 by Shrinava last updated on 29/Apr/22

1. 2(√y) dx = dy 2. (x^2 + y^2 )dx = 2xydy 3. xdx + (1/y) dy = 0 4. dy = 3x^2 dx 5. 2y^2 dx + x(1 + y^2 ) dy = 0 6. 4x^3 dx + dy = 0

$$\mathrm{1}.\:\mathrm{2}\sqrt{\mathrm{y}}\:\mathrm{dx}\:=\:\mathrm{dy} \\ $$$$\mathrm{2}.\:\left(\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} \right)\mathrm{dx}\:=\:\mathrm{2xydy} \\ $$$$\mathrm{3}.\:\mathrm{xdx}\:+\:\frac{\mathrm{1}}{\mathrm{y}}\:\mathrm{dy}\:=\:\mathrm{0} \\ $$$$\mathrm{4}.\:\mathrm{dy}\:=\:\mathrm{3x}^{\mathrm{2}} \:\mathrm{dx} \\ $$$$\mathrm{5}.\:\mathrm{2y}^{\mathrm{2}} \mathrm{dx}\:+\:\mathrm{x}\left(\mathrm{1}\:+\:\mathrm{y}^{\mathrm{2}} \right)\:\mathrm{dy}\:=\:\mathrm{0} \\ $$$$\mathrm{6}.\:\mathrm{4x}^{\mathrm{3}} \mathrm{dx}\:+\:\mathrm{dy}\:=\:\mathrm{0} \\ $$

Commented by mkam last updated on 29/Apr/22

(1) dx = (dy/(2(√y))) ⇒ x = (√y) + C ⇒ y = ( x+c )^2 (2) (dy/dx) = ((x^2 +y^2 )/(2xy)) ⇒ v + x (dv/dx) = ((x^2 (1+v^2 ))/(2x^2 v)) ⇒ x(dv/dx) = ((1+v^2 )/(2v)) − v ⇒ x (dv/(dx ))= ((1−v^2 )/(2v)) ⇒ (dx/x) = ((2v)/(−(v^2 −1)))dv ⇒ lnx + lnc = − 2ln∣v^2 −1∣ ⇒ ln(xc)= ln∣(1/((v^2 −1)))∣ ⇒ xc = (x^2 /(y^2 −x^2 )) (3)xdx+(1/y)dy=0 ⇒ (x^2 /2) + lny = C (4)dy = 3x^2 dx ⇒ y = x^3 + C (5)2y^2 dx+x(1+y^2 )dy = 0 ⇒ ((2dx)/x) + ((1/y^2 ) + 1 )dy = 0 ⇒ 2lnx −(1/y) + y = C (6)4x^3 dx + dy = 0 ⇒ x^4 +y = C aldolaimy

$$\left(\mathrm{1}\right)\:{dx}\:=\:\frac{{dy}}{\mathrm{2}\sqrt{{y}}}\:\:\Rightarrow\:{x}\:=\:\sqrt{{y}}\:+\:{C}\:\Rightarrow\:{y}\:=\:\left(\:{x}+{c}\:\right)^{\mathrm{2}} \\ $$$$ \\ $$$$\left(\mathrm{2}\right)\:\frac{{dy}}{{dx}}\:=\:\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{\mathrm{2}{xy}}\:\Rightarrow\:{v}\:+\:{x}\:\frac{{dv}}{{dx}}\:=\:\frac{{x}^{\mathrm{2}} \left(\mathrm{1}+{v}^{\mathrm{2}} \right)}{\mathrm{2}{x}^{\mathrm{2}} \:{v}}\:\Rightarrow\:\:{x}\frac{{dv}}{{dx}}\:=\:\frac{\mathrm{1}+{v}^{\mathrm{2}} }{\mathrm{2}{v}}\:−\:{v} \\ $$$$ \\ $$$$\Rightarrow\:{x}\:\frac{{dv}}{{dx}\:}=\:\frac{\mathrm{1}−{v}^{\mathrm{2}} }{\mathrm{2}{v}}\:\Rightarrow\:\frac{{dx}}{{x}}\:=\:\frac{\mathrm{2}{v}}{−\left({v}^{\mathrm{2}} −\mathrm{1}\right)}{dv}\:\Rightarrow\:{lnx}\:+\:{lnc}\:=\:−\:\mathrm{2}{ln}\mid{v}^{\mathrm{2}} −\mathrm{1}\mid \\ $$$$ \\ $$$$\Rightarrow\:{ln}\left({xc}\right)=\:{ln}\mid\frac{\mathrm{1}}{\left({v}^{\mathrm{2}} −\mathrm{1}\right)}\mid\:\Rightarrow\:{xc}\:=\:\frac{{x}^{\mathrm{2}} }{{y}^{\mathrm{2}} −{x}^{\mathrm{2}} } \\ $$$$ \\ $$$$\left(\mathrm{3}\right){xdx}+\frac{\mathrm{1}}{{y}}{dy}=\mathrm{0}\:\Rightarrow\:\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:+\:{lny}\:=\:{C} \\ $$$$ \\ $$$$\left(\mathrm{4}\right){dy}\:=\:\mathrm{3}{x}^{\mathrm{2}} {dx}\:\Rightarrow\:{y}\:=\:{x}^{\mathrm{3}} +\:{C} \\ $$$$ \\ $$$$\left(\mathrm{5}\right)\mathrm{2}{y}^{\mathrm{2}} {dx}+{x}\left(\mathrm{1}+{y}^{\mathrm{2}} \right){dy}\:=\:\mathrm{0}\:\Rightarrow\:\frac{\mathrm{2}{dx}}{{x}}\:+\:\left(\frac{\mathrm{1}}{{y}^{\mathrm{2}} }\:+\:\mathrm{1}\:\right){dy}\:=\:\mathrm{0}\: \\ $$$$\Rightarrow\:\mathrm{2}{lnx}\:−\frac{\mathrm{1}}{{y}}\:+\:{y}\:=\:{C} \\ $$$$ \\ $$$$\left(\mathrm{6}\right)\mathrm{4}{x}^{\mathrm{3}} {dx}\:+\:{dy}\:=\:\mathrm{0}\:\Rightarrow\:{x}^{\mathrm{4}} +{y}\:=\:{C} \\ $$$$ \\ $$$${aldolaimy} \\ $$$$ \\ $$

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