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Question Number 168613 by LEKOUMA last updated on 14/Apr/22

Resolve 1) x(dy/dx)−y=y^3 2) (x−y)ydx−x^2 dy=0 3) (2x−y)dx+(4x−2y+3)dy=0

$${Resolve}\: \\ $$$$\left.\mathrm{1}\right)\:{x}\frac{{dy}}{{dx}}−{y}={y}^{\mathrm{3}} \\ $$$$\left.\mathrm{2}\right)\:\left({x}−{y}\right){ydx}−{x}^{\mathrm{2}} {dy}=\mathrm{0} \\ $$$$\left.\mathrm{3}\right)\:\left(\mathrm{2}{x}−{y}\right){dx}+\left(\mathrm{4}{x}−\mathrm{2}{y}+\mathrm{3}\right){dy}=\mathrm{0} \\ $$

Commented by Mastermind last updated on 14/Apr/22

Solution 1 x(dy/dx)−y=y^3 ⇒ (1/y^3 )(dy/dx)−(1/(xy^2 ))=(1/x) .........(1) put z=−(1/y^2 ) , so that (dz/dx)=2(1/y^3 )(dy/dx) ⇒(1/2)(dz/dx)=(1/y^3 )(dy/dx) from (1), we have (1/2)(dz/dx)−(z/x)=(1/x) multiply all through by 2, then (dz/dx)−((2z)/x)=(2/x) ........(2) Now, IF = e^(∫−(2/x)) ⇒e^(ln(x^(−2) )) ⇒ x^(−2) =(1/x^2 ) multiply both sides by IF we have, (1/x^2 )((dz/dx)−((2z)/x))=(2/x^3 ) Now, (d/dx)(z∙(1/x^2 ))=(2/x^3 ) d(z∙(1/x^2 ))=(2/x^3 )dx On integrating, we get (z/x^2 )=−(1/x^2 )+c ⇒z=−1+cx^2 putting z=−(1/y^2 ) We have, −(1/y^2 )=−1+cx^2 ⇒ (1/y^2 )=1−cx^2 Resolved Mastermind

$$ \\ $$$${Solution}\:\mathrm{1} \\ $$$${x}\frac{{dy}}{{dx}}−{y}={y}^{\mathrm{3}} \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{{y}^{\mathrm{3}} }\frac{{dy}}{{dx}}−\frac{\mathrm{1}}{{xy}^{\mathrm{2}} }=\frac{\mathrm{1}}{{x}}\:.........\left(\mathrm{1}\right) \\ $$$${put}\:{z}=−\frac{\mathrm{1}}{{y}^{\mathrm{2}} }\:,\:{so}\:{that}\:\frac{{dz}}{{dx}}=\mathrm{2}\frac{\mathrm{1}}{{y}^{\mathrm{3}} }\frac{{dy}}{{dx}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}\frac{{dz}}{{dx}}=\frac{\mathrm{1}}{{y}^{\mathrm{3}} }\frac{{dy}}{{dx}} \\ $$$${from}\:\left(\mathrm{1}\right),\:{we}\:{have} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\frac{{dz}}{{dx}}−\frac{{z}}{{x}}=\frac{\mathrm{1}}{{x}} \\ $$$${multiply}\:{all}\:{through}\:{by}\:\mathrm{2},\:{then} \\ $$$$\frac{{dz}}{{dx}}−\frac{\mathrm{2}{z}}{{x}}=\frac{\mathrm{2}}{{x}}\:........\left(\mathrm{2}\right) \\ $$$${Now},\:{IF}\:=\:{e}^{\int−\frac{\mathrm{2}}{{x}}} \Rightarrow{e}^{{ln}\left({x}^{−\mathrm{2}} \right)} \\ $$$$\Rightarrow\:{x}^{−\mathrm{2}} =\frac{\mathrm{1}}{{x}^{\mathrm{2}} } \\ $$$${multiply}\:{both}\:{sides}\:{by}\:{IF} \\ $$$${we}\:{have},\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\left(\frac{{dz}}{{dx}}−\frac{\mathrm{2}{z}}{{x}}\right)=\frac{\mathrm{2}}{{x}^{\mathrm{3}} } \\ $$$${Now},\:\frac{{d}}{{dx}}\left({z}\centerdot\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)=\frac{\mathrm{2}}{{x}^{\mathrm{3}} } \\ $$$${d}\left({z}\centerdot\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)=\frac{\mathrm{2}}{{x}^{\mathrm{3}} }{dx} \\ $$$${On}\:{integrating},\:{we}\:{get} \\ $$$$\frac{{z}}{{x}^{\mathrm{2}} }=−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+{c} \\ $$$$\Rightarrow{z}=−\mathrm{1}+{cx}^{\mathrm{2}} \\ $$$${putting}\:{z}=−\frac{\mathrm{1}}{{y}^{\mathrm{2}} } \\ $$$${We}\:{have},\: \\ $$$$−\frac{\mathrm{1}}{{y}^{\mathrm{2}} }=−\mathrm{1}+{cx}^{\mathrm{2}} \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{{y}^{\mathrm{2}} }=\mathrm{1}−{cx}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{Resolved} \\ $$$$ \\ $$$${Mastermind} \\ $$

Answered by mr W last updated on 15/Apr/22

(1) (dy/(y(1+y^2 )))=(dx/x) ((1/y)−(y/(1+y^2 )))dy=(dx/x) ln y−(1/2)ln (1+y^2 )=ln x+C ⇒(y^2 /( 1+y^2 ))=Cx^2 or ⇒(1−Cx^2 )(1+y^2 )=1 or ⇒(1/(1+y^2 ))=1−Cx^2

$$\left(\mathrm{1}\right) \\ $$$$\frac{{dy}}{{y}\left(\mathrm{1}+{y}^{\mathrm{2}} \right)}=\frac{{dx}}{{x}} \\ $$$$\left(\frac{\mathrm{1}}{{y}}−\frac{{y}}{\mathrm{1}+{y}^{\mathrm{2}} }\right){dy}=\frac{{dx}}{{x}} \\ $$$$\mathrm{ln}\:{y}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left(\mathrm{1}+{y}^{\mathrm{2}} \right)=\mathrm{ln}\:{x}+{C} \\ $$$$\Rightarrow\frac{{y}^{\mathrm{2}} }{\:\mathrm{1}+{y}^{\mathrm{2}} }={Cx}^{\mathrm{2}} \\ $$$${or} \\ $$$$\Rightarrow\left(\mathrm{1}−{Cx}^{\mathrm{2}} \right)\left(\mathrm{1}+{y}^{\mathrm{2}} \right)=\mathrm{1} \\ $$$${or} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{1}+{y}^{\mathrm{2}} }=\mathrm{1}−{Cx}^{\mathrm{2}} \\ $$

Answered by mr W last updated on 15/Apr/22

(2) (dy/dx)=(((x−y)y)/x^2 ) y=xt (dy/dx)=t+x(dt/dx) t+x(dt/dx)=t(1−t) x(dt/dx)=−t^2 (dt/t^2 )=−(dx/x) ∫(dt/t^2 )=−∫(dx/x) −(1/t)=−ln x−C (x/y)=ln x+C ⇒y=(x/(ln x+C))

$$\left(\mathrm{2}\right) \\ $$$$\frac{{dy}}{{dx}}=\frac{\left({x}−{y}\right){y}}{{x}^{\mathrm{2}} } \\ $$$${y}={xt} \\ $$$$\frac{{dy}}{{dx}}={t}+{x}\frac{{dt}}{{dx}} \\ $$$${t}+{x}\frac{{dt}}{{dx}}={t}\left(\mathrm{1}−{t}\right) \\ $$$${x}\frac{{dt}}{{dx}}=−{t}^{\mathrm{2}} \\ $$$$\frac{{dt}}{{t}^{\mathrm{2}} }=−\frac{{dx}}{{x}} \\ $$$$\int\frac{{dt}}{{t}^{\mathrm{2}} }=−\int\frac{{dx}}{{x}} \\ $$$$−\frac{\mathrm{1}}{{t}}=−\mathrm{ln}\:{x}−{C} \\ $$$$\frac{{x}}{{y}}=\mathrm{ln}\:{x}+{C} \\ $$$$\Rightarrow{y}=\frac{{x}}{\mathrm{ln}\:{x}+{C}} \\ $$

Answered by mr W last updated on 15/Apr/22

(3) (dy/dx)=((−2x+y)/(4x−2y+3)) let u=−2x+y (du/dx)=−2+(dy/dx) 2+(du/dx)=(u/(−2u+3)) (du/dx)=((5u−6)/(−2u+3)) (((2u−3)du)/(5u−6))=−dx (((10u−12−3)du)/(5u−6))=−5dx (2−(3/(5u−6)))du=−5dx 2u−(3/5)ln (5u−6)=−5x+C −4x+2y−(3/5)ln (−10x+5y−6)=−5x+C ⇒x+2y−(3/5)ln (−10x+5y−6)=C

$$\left(\mathrm{3}\right) \\ $$$$\frac{{dy}}{{dx}}=\frac{−\mathrm{2}{x}+{y}}{\mathrm{4}{x}−\mathrm{2}{y}+\mathrm{3}} \\ $$$${let}\:{u}=−\mathrm{2}{x}+{y} \\ $$$$\frac{{du}}{{dx}}=−\mathrm{2}+\frac{{dy}}{{dx}} \\ $$$$\mathrm{2}+\frac{{du}}{{dx}}=\frac{{u}}{−\mathrm{2}{u}+\mathrm{3}} \\ $$$$\frac{{du}}{{dx}}=\frac{\mathrm{5}{u}−\mathrm{6}}{−\mathrm{2}{u}+\mathrm{3}} \\ $$$$\frac{\left(\mathrm{2}{u}−\mathrm{3}\right){du}}{\mathrm{5}{u}−\mathrm{6}}=−{dx} \\ $$$$\frac{\left(\mathrm{10}{u}−\mathrm{12}−\mathrm{3}\right){du}}{\mathrm{5}{u}−\mathrm{6}}=−\mathrm{5}{dx} \\ $$$$\left(\mathrm{2}−\frac{\mathrm{3}}{\mathrm{5}{u}−\mathrm{6}}\right){du}=−\mathrm{5}{dx} \\ $$$$\mathrm{2}{u}−\frac{\mathrm{3}}{\mathrm{5}}\mathrm{ln}\:\left(\mathrm{5}{u}−\mathrm{6}\right)=−\mathrm{5}{x}+{C} \\ $$$$−\mathrm{4}{x}+\mathrm{2}{y}−\frac{\mathrm{3}}{\mathrm{5}}\mathrm{ln}\:\left(−\mathrm{10}{x}+\mathrm{5}{y}−\mathrm{6}\right)=−\mathrm{5}{x}+{C} \\ $$$$\Rightarrow{x}+\mathrm{2}{y}−\frac{\mathrm{3}}{\mathrm{5}}\mathrm{ln}\:\left(−\mathrm{10}{x}+\mathrm{5}{y}−\mathrm{6}\right)={C} \\ $$

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