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Question Number 116136 by bemath last updated on 01/Oct/20

y′ =(y/x) +((2x^3 cos (x^2 ))/y) where y((√π)) = 0

$$\mathrm{y}'\:=\frac{\mathrm{y}}{\mathrm{x}}\:+\frac{\mathrm{2x}^{\mathrm{3}} \:\mathrm{cos}\:\left(\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{y}} \\ $$$$\mathrm{where}\:\mathrm{y}\left(\sqrt{\pi}\right)\:=\:\mathrm{0} \\ $$

Answered by mindispower last updated on 01/Oct/20

(y/x)=z y′=z+xz′ ⇒z+xz′=z+((2x^3 cos(x^2 ))/(xz)) ⇒z(xz′)=2x^2 cos(x^2 ) ⇒zz′=2xcos(x^2 ) ⇒∫zdz=∫2xcos(x^2 )dx ⇔(z^2 /2)=sin(x^2 )+c z= +_− 2(√(sin(x^2 )+c)) y= +_− z(√(sin(x^2 )+c)) y((√π))=+_− 2(√π).(√(sin(π)+c))=0⇒c=0 y(x)= +_− 2(√(sin(x^2 )))

$$\frac{{y}}{{x}}={z} \\ $$$${y}'={z}+{xz}' \\ $$$$\Rightarrow{z}+{xz}'={z}+\frac{\mathrm{2}{x}^{\mathrm{3}} {cos}\left({x}^{\mathrm{2}} \right)}{{xz}} \\ $$$$\Rightarrow{z}\left({xz}'\right)=\mathrm{2}{x}^{\mathrm{2}} {cos}\left({x}^{\mathrm{2}} \right) \\ $$$$\Rightarrow{zz}'=\mathrm{2}{xcos}\left({x}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\int{zdz}=\int\mathrm{2}{xcos}\left({x}^{\mathrm{2}} \right){dx} \\ $$$$\Leftrightarrow\frac{{z}^{\mathrm{2}} }{\mathrm{2}}={sin}\left({x}^{\mathrm{2}} \right)+{c} \\ $$$${z}=\:\:\underset{−} {+}\mathrm{2}\sqrt{{sin}\left({x}^{\mathrm{2}} \right)+{c}} \\ $$$${y}=\:\:\underset{−} {+}{z}\sqrt{{sin}\left({x}^{\mathrm{2}} \right)+{c}} \\ $$$${y}\left(\sqrt{\pi}\right)=\underset{−} {+}\mathrm{2}\sqrt{\pi}.\sqrt{{sin}\left(\pi\right)+{c}}=\mathrm{0}\Rightarrow{c}=\mathrm{0} \\ $$$${y}\left({x}\right)=\:\:\underset{−} {+}\mathrm{2}\sqrt{{sin}\left({x}^{\mathrm{2}} \right)} \\ $$$$ \\ $$$$ \\ $$

Answered by TANMAY PANACEA last updated on 01/Oct/20

(dy/dx)−(y/x)=((2x^3 cosx^2 )/y) ((xdy−ydx)/(xdx×x^2 ))=((2cosx^2 )/(((y/x)))) d((y/x)).(y/x)=cosx^2 .dx^2 intregating ∫((y/x))d((y/x))=∫cosx^2 .dx^2 (1/2)((y/x))^2 =sinx^2 +c

$$\frac{{dy}}{{dx}}−\frac{{y}}{{x}}=\frac{\mathrm{2}{x}^{\mathrm{3}} {cosx}^{\mathrm{2}} }{{y}} \\ $$$$\frac{{xdy}−{ydx}}{{xdx}×{x}^{\mathrm{2}} }=\frac{\mathrm{2}{cosx}^{\mathrm{2}} }{\left(\frac{{y}}{{x}}\right)} \\ $$$${d}\left(\frac{{y}}{{x}}\right).\frac{{y}}{{x}}={cosx}^{\mathrm{2}} .{dx}^{\mathrm{2}} \\ $$$${intregating} \\ $$$$\int\left(\frac{{y}}{{x}}\right){d}\left(\frac{{y}}{{x}}\right)=\int{cosx}^{\mathrm{2}} .{dx}^{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{y}}{{x}}\right)^{\mathrm{2}} ={sinx}^{\mathrm{2}} +{c} \\ $$

Answered by mathmax by abdo last updated on 01/Oct/20

y^′ =(y/x) +((2x^3 cos(x^2 ))/y) let (y/x)=z ⇒y =xz ⇒y^(′ ) =z+xz^′ (e)⇒z+xz^′ =z +((2x^2 cos(x^2 ))/z) ⇒xzz^′ =2x^2 cos(x^2 ) ⇒ zz^′ =2xcos(x^2 ) ⇒∫ zz^′ dx =2∫ xcos(x^2 )dx =sin(x^2 )+c ⇒ (1/2)z^2 =sin(x^2 )+c ⇒z^2 =2sin(x^2 ) +2c ⇒z =+^− (√(2sin(x^2 )+λ))

$$\mathrm{y}^{'} \:=\frac{\mathrm{y}}{\mathrm{x}}\:+\frac{\mathrm{2x}^{\mathrm{3}} \mathrm{cos}\left(\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{y}}\:\:\mathrm{let}\:\frac{\mathrm{y}}{\mathrm{x}}=\mathrm{z}\:\Rightarrow\mathrm{y}\:=\mathrm{xz}\:\Rightarrow\mathrm{y}^{'\:} =\mathrm{z}+\mathrm{xz}^{'} \\ $$$$\left(\mathrm{e}\right)\Rightarrow\mathrm{z}+\mathrm{xz}^{'} \:=\mathrm{z}\:+\frac{\mathrm{2x}^{\mathrm{2}} \:\mathrm{cos}\left(\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{z}}\:\Rightarrow\mathrm{xzz}^{'} \:=\mathrm{2x}^{\mathrm{2}} \mathrm{cos}\left(\mathrm{x}^{\mathrm{2}} \right)\:\Rightarrow \\ $$$$\mathrm{zz}^{'} \:=\mathrm{2xcos}\left(\mathrm{x}^{\mathrm{2}} \right)\:\Rightarrow\int\:\mathrm{zz}^{'} \mathrm{dx}\:=\mathrm{2}\int\:\mathrm{xcos}\left(\mathrm{x}^{\mathrm{2}} \right)\mathrm{dx}\:=\mathrm{sin}\left(\mathrm{x}^{\mathrm{2}} \right)+\mathrm{c}\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\mathrm{z}^{\mathrm{2}} \:=\mathrm{sin}\left(\mathrm{x}^{\mathrm{2}} \right)+\mathrm{c}\:\Rightarrow\mathrm{z}^{\mathrm{2}} \:=\mathrm{2sin}\left(\mathrm{x}^{\mathrm{2}} \right)\:+\mathrm{2c}\:\Rightarrow\mathrm{z}\:=\overset{−} {+}\sqrt{\mathrm{2sin}\left(\mathrm{x}^{\mathrm{2}} \right)+\lambda} \\ $$$$ \\ $$

Commented by mathmax by abdo last updated on 01/Oct/20

⇒y =+^− x(√(2sin(x^2 )+λ)) y((√π)) =0 ⇒+^− (√π)(√(0+λ))=0⇒λ =0 ⇒y =+^− (√(2sin(x^2 )))

$$\Rightarrow\mathrm{y}\:=\overset{−} {+}\mathrm{x}\sqrt{\mathrm{2sin}\left(\mathrm{x}^{\mathrm{2}} \right)+\lambda}\: \\ $$$$\mathrm{y}\left(\sqrt{\pi}\right)\:=\mathrm{0}\:\Rightarrow\overset{−} {+}\sqrt{\pi}\sqrt{\mathrm{0}+\lambda}=\mathrm{0}\Rightarrow\lambda\:=\mathrm{0}\:\Rightarrow\mathrm{y}\:=\overset{−} {+}\sqrt{\mathrm{2sin}\left(\mathrm{x}^{\mathrm{2}} \right)} \\ $$

Commented by mathmax by abdo last updated on 02/Oct/20

y =+^− x(√(2sin(x^2 )))

$$\mathrm{y}\:=\overset{−} {+}\mathrm{x}\sqrt{\mathrm{2sin}\left(\mathrm{x}^{\mathrm{2}} \right)} \\ $$

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