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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
y=yx+2x3cos(x2)ywherey(π)=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 )))
yx=zy=z+xzz+xz=z+2x3cos(x2)xzz(xz)=2x2cos(x2)zz=2xcos(x2)zdz=2xcos(x2)dxz22=sin(x2)+cz=+2sin(x2)+cy=+zsin(x2)+cy(π)=+2π.sin(π)+c=0c=0y(x)=+2sin(x2)
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
dydxyx=2x3cosx2yxdyydxxdx×x2=2cosx2(yx)d(yx).yx=cosx2.dx2intregating(yx)d(yx)=cosx2.dx212(yx)2=sinx2+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))
y=yx+2x3cos(x2)yletyx=zy=xzy=z+xz(e)z+xz=z+2x2cos(x2)zxzz=2x2cos(x2)zz=2xcos(x2)zzdx=2xcos(x2)dx=sin(x2)+c12z2=sin(x2)+cz2=2sin(x2)+2cz=+2sin(x2)+λ
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 )))
y=+x2sin(x2)+λy(π)=0+π0+λ=0λ=0y=+2sin(x2)
Commented by mathmax by abdo last updated on 02/Oct/20
y =+^− x(√(2sin(x^2 )))
y=+x2sin(x2)

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