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solve-the-de-x-2-1-y-xy-x-2-x-




Question Number 65402 by mathmax by abdo last updated on 29/Jul/19
solve the (de)    (√(x^2 +1))y^(′′)    −xy^′  =x^2 −x
solvethe(de)x2+1yxy=x2x
Commented by mathmax by abdo last updated on 30/Jul/19
let y^′ =z    (e) ⇒(√(x^2 +1))z^′ −xz =x^2 −x  (he) ⇒(√(x^2 +1))z^′ −xz =0 ⇒(√(x^2  +1))z^′ =xz ⇒(z^′ /z) =(x/( (√(x^2  +1)))) ⇒  ln∣z∣ =(√(x^2  +1 )) +c ⇒z =k e^(√(1+x^2 ))   mvc method give  z^′  =k^′ e^(√(1+x^2 ))  +k  (x/( (√(1+x^2 ))))e^(√(1+x^2 ))   (e) ⇒(√(1+x^2 ))( k^′  +((kx)/( (√(1+x^2 )))))e^(√(1+x^2 )) −xke^(√(1+x^2 ))  =x^2 −x ⇒  (√(1+x^2  ))k^′  e^(√(1+x^2 ))  +kxe^(√(1+x^2 )) −xk e^(√(1+x^2 )) =x^2 −x ⇒  k^′  =((x^2 −x)/( (√(1+x^2 )))) e^(−(√(1+x^2 )))   ⇒k(x) =∫   ((x^2 −x)/( (√(1+x^2 ))))e^(−(√(1+x^2 )))  dx +c ⇒  z(x) =( ∫ ((x^2 −x)/( (√(1+x^2 ))))e^(−(√(1+x^2 ))) dx +c)e^(√(1+x^2 ))   =e^(√(1+x^2 ))  ∫   ((x^2 −x)/( (√(1+x^2 ))))e^(−(√(1+x^2 ))) dx  +c e^(√(1+x^2 ))   y^′  =z(x) =⇒y(x) =∫^x z(t)dt +λ  =∫^x  {e^(√(1+t^2 ))  ∫^t  ((u^2 −u)/( (√(1+u^2 ))))e^(−(√(1+u^2 ))) du  +c e^(√(1+t^2 )) }dt +λ
lety=z(e)x2+1zxz=x2x(he)x2+1zxz=0x2+1z=xzzz=xx2+1lnz=x2+1+cz=ke1+x2mvcmethodgivez=ke1+x2+kx1+x2e1+x2(e)1+x2(k+kx1+x2)e1+x2xke1+x2=x2x1+x2ke1+x2+kxe1+x2xke1+x2=x2xk=x2x1+x2e1+x2k(x)=x2x1+x2e1+x2dx+cz(x)=(x2x1+x2e1+x2dx+c)e1+x2=e1+x2x2x1+x2e1+x2dx+ce1+x2y=z(x)=⇒y(x)=xz(t)dt+λ=x{e1+t2tu2u1+u2e1+u2du+ce1+t2}dt+λ

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