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Question Number 105283 by bemath last updated on 27/Jul/20

(1) (dy/dx) = ((2xy)/(4x^2 −y^3 )) (2) (dy/dx) = ((sin x+cos x)/(y(2ln y + 1)))

$$\left(\mathrm{1}\right)\:\frac{{dy}}{{dx}}\:=\:\frac{\mathrm{2}{xy}}{\mathrm{4}{x}^{\mathrm{2}} −{y}^{\mathrm{3}} } \\ $$$$\left(\mathrm{2}\right)\:\frac{{dy}}{{dx}}\:=\:\frac{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}{{y}\left(\mathrm{2ln}\:{y}\:+\:\mathrm{1}\right)} \\ $$

Answered by bobhans last updated on 27/Jul/20

(2) y(2ln y+1) dy = (sin x+cos x) dx ∫ (2y ln y +y)dy = ∫ (sin x+cos x)dx (1/2)y^2 +y^2 ln y −∫y dy = sin x−cos x +C (1/2)y^2 +y^2 ln y−(1/2)y^2 =sin x−cos x+C →y^2 ln y = sin x−cos x + C ▷

$$\left(\mathrm{2}\right)\:{y}\left(\mathrm{2ln}\:{y}+\mathrm{1}\right)\:{dy}\:=\:\left(\mathrm{sin}\:{x}+\mathrm{cos}\:{x}\right)\:{dx} \\ $$$$\int\:\left(\mathrm{2}{y}\:\mathrm{ln}\:{y}\:+{y}\right){dy}\:=\:\int\:\left(\mathrm{sin}\:{x}+\mathrm{cos}\:{x}\right){dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{y}^{\mathrm{2}} +{y}^{\mathrm{2}} \mathrm{ln}\:{y}\:−\int{y}\:{dy}\:=\:\mathrm{sin}\:{x}−\mathrm{cos}\:{x}\:+{C} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{y}^{\mathrm{2}} +{y}^{\mathrm{2}} \mathrm{ln}\:{y}−\frac{\mathrm{1}}{\mathrm{2}}{y}^{\mathrm{2}} =\mathrm{sin}\:{x}−\mathrm{cos}\:{x}+{C} \\ $$$$\rightarrow{y}^{\mathrm{2}} \:\mathrm{ln}\:{y}\:=\:\mathrm{sin}\:{x}−\mathrm{cos}\:{x}\:+\:{C}\:\triangleright \\ $$

Answered by Dwaipayan Shikari last updated on 27/Jul/20

(dy/dx)=((sinx+cosx)/(y(2logy+1))) ∫y(2logy+1)dy=∫sinx+cosx dx y^2 logy−(y^2 /2)+(y^2 /2)=−cosx+sinx+C y^2 logy=−cosx+sinx+C y=e^((−cosx+sinx)/y^2 )

$$\frac{{dy}}{{dx}}=\frac{{sinx}+{cosx}}{{y}\left(\mathrm{2}{logy}+\mathrm{1}\right)} \\ $$$$\int{y}\left(\mathrm{2}{logy}+\mathrm{1}\right){dy}=\int{sinx}+{cosx}\:{dx} \\ $$$${y}^{\mathrm{2}} {logy}−\frac{{y}^{\mathrm{2}} }{\mathrm{2}}+\frac{{y}^{\mathrm{2}} }{\mathrm{2}}=−{cosx}+{sinx}+{C} \\ $$$${y}^{\mathrm{2}} {logy}=−{cosx}+{sinx}+{C} \\ $$$${y}={e}^{\frac{−{cosx}+{sinx}}{{y}^{\mathrm{2}} }} \\ $$

Answered by bobhans last updated on 27/Jul/20

(1) (y^3 −4x^2 ) dy + 2xy dx = 0 (∂N/∂x) = −8x ; (∂M/∂y) = 2x ⇔ (∂N/∂x) ≠ (∂M/∂y) non exact . by observe (((∂N/∂x)−(∂M/∂y))/M) = ((−10x)/(2xy)) = −(5/y) integrating factor u(y)=e^(∫ −(5/y) dy) u(y) = e^(−5ln y) = y^(−5) we can multiply original equation by y^(−5) ⇒(y^(−2) −4x^2 y^(−5) )dy +2xy^(−4) dx = 0 { (((∂F/∂x) = M(x,y) = 2xy^(−4) )),(((∂F/∂x) = N(x,y) = y^(−2) −4x^2 y^(−5) )) :} F(x,y) = ∫(2xy^(−4) )dx + g(y) F(x,y) = x^2 y^(−4) + g(y) ⇔g′(y) = N(x,y)−(∂/∂y) (x^2 y^(−4) ) g′(y) = y^(−2) −4x^2 y^(−5) +4x^2 y^(−5) g′(y) = y^(−2) →g(y) = ∫y^(−2) dy g(y) = −y^(−1) . Therefore we get solution ∴ x^2 y^(−4) −y^(−1) = C . ▷

$$\left(\mathrm{1}\right)\:\left({y}^{\mathrm{3}} −\mathrm{4}{x}^{\mathrm{2}} \right)\:{dy}\:+\:\mathrm{2}{xy}\:{dx}\:=\:\mathrm{0} \\ $$$$\frac{\partial{N}}{\partial{x}}\:=\:−\mathrm{8}{x}\:;\:\frac{\partial{M}}{\partial{y}}\:=\:\mathrm{2}{x}\:\Leftrightarrow\:\frac{\partial{N}}{\partial{x}}\:\neq\:\frac{\partial{M}}{\partial{y}} \\ $$$${non}\:{exact}\:. \\ $$$${by}\:{observe}\:\frac{\frac{\partial{N}}{\partial{x}}−\frac{\partial{M}}{\partial{y}}}{{M}}\:=\:\frac{−\mathrm{10}{x}}{\mathrm{2}{xy}}\:=\:−\frac{\mathrm{5}}{{y}} \\ $$$${integrating}\:{factor}\:{u}\left({y}\right)={e}^{\int\:−\frac{\mathrm{5}}{{y}}\:{dy}} \\ $$$${u}\left({y}\right)\:=\:{e}^{−\mathrm{5ln}\:{y}} \:=\:{y}^{−\mathrm{5}} \\ $$$${we}\:{can}\:{multiply}\:{original}\:{equation}\:{by}\:{y}^{−\mathrm{5}} \\ $$$$\Rightarrow\left({y}^{−\mathrm{2}} −\mathrm{4}{x}^{\mathrm{2}} {y}^{−\mathrm{5}} \right){dy}\:+\mathrm{2}{xy}^{−\mathrm{4}} \:{dx}\:=\:\mathrm{0} \\ $$$$\begin{cases}{\frac{\partial{F}}{\partial{x}}\:=\:{M}\left({x},{y}\right)\:=\:\mathrm{2}{xy}^{−\mathrm{4}} }\\{\frac{\partial{F}}{\partial{x}}\:=\:{N}\left({x},{y}\right)\:=\:{y}^{−\mathrm{2}} −\mathrm{4}{x}^{\mathrm{2}} {y}^{−\mathrm{5}} }\end{cases} \\ $$$${F}\left({x},{y}\right)\:=\:\int\left(\mathrm{2}{xy}^{−\mathrm{4}} \right){dx}\:+\:{g}\left({y}\right) \\ $$$${F}\left({x},{y}\right)\:=\:{x}^{\mathrm{2}} {y}^{−\mathrm{4}} \:+\:{g}\left({y}\right) \\ $$$$\Leftrightarrow{g}'\left({y}\right)\:=\:{N}\left({x},{y}\right)−\frac{\partial}{\partial{y}}\:\left({x}^{\mathrm{2}} {y}^{−\mathrm{4}} \right) \\ $$$${g}'\left({y}\right)\:=\:{y}^{−\mathrm{2}} −\mathrm{4}{x}^{\mathrm{2}} {y}^{−\mathrm{5}} +\mathrm{4}{x}^{\mathrm{2}} {y}^{−\mathrm{5}} \\ $$$${g}'\left({y}\right)\:=\:{y}^{−\mathrm{2}} \:\rightarrow{g}\left({y}\right)\:=\:\int{y}^{−\mathrm{2}} {dy} \\ $$$${g}\left({y}\right)\:=\:−{y}^{−\mathrm{1}} .\:{Therefore}\:{we}\:{get}\:{solution} \\ $$$$\therefore\:{x}^{\mathrm{2}} {y}^{−\mathrm{4}} −{y}^{−\mathrm{1}} \:=\:{C}\:.\:\triangleright \\ $$$$ \\ $$

Commented by bemath last updated on 27/Jul/20

very...cooll thanks

$${very}...{cooll}\:{thanks}\: \\ $$

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