Menu Close

To-solve-x-3-x-c-Let-y-x-p-x-3-x-c-dy-dx-x-3-x-c-3x-2-1-x-p-let-dy-dx-mx-3x-2-1-x-p-mx-3x-3-3px-2-m-1-x-p-0-3-x-c-3px-2-m-1-x-p-0-3px-2-m-2-x-3c-p-0-and-since

Tinku Tara 0 Comments
Pin




Question Number 130089 by ajfour last updated on 22/Jan/21
To solve x^3 =x+c Let y=(x−p)(x^3 −x−c) (dy/dx)=(x^3 −x−c)+(3x^2 −1)(x−p) let (dy/dx)=mx ⇒ (3x^2 −1)(x−p)=mx 3x^3 −3px^2 −(m+1)x+p=0 3(x+c)−3px^2 −(m+1)x+p=0 3px^2 +(m−2)x−(3c+p)=0 and since x^3 =x+c (m−2)x^2 +(2p−3c)x+3cp=0 [9cp^2 +(3c+p)(m−2)]x +[3cp(m−2)+(3c+p)(2p−3c)] =0 x=−(((3c+p)(2p−3c)+3cp(m−2))/(9cp^2 +(3c+p)(m−2))) (m−2)x^2 +(2p−3c)x+(3cp)=0 Now choosing suitable ′p′ value we find ′m′; and then x=f(p,m) ★
Tosolvex3=x+cLety=(x−p)(x3−x−c)dydx=(x3−x−c)+(3x2−1)(x−p)letdydx=mx⇒(3x2−1)(x−p)=mx3x3−3px2−(m+1)x+p=03(x+c)−3px2−(m+1)x+p=03px2+(m−2)x−(3c+p)=0andsincex3=x+c(m−2)x2+(2p−3c)x+3cp=0[9cp2+(3c+p)(m−2)]x+[3cp(m−2)+(3c+p)(2p−3c)]=0x=−(3c+p)(2p−3c)+3cp(m−2)9cp2+(3c+p)(m−2)(m−2)x2+(2p−3c)x+(3cp)=0Nowchoosingsuitable′p′valuewefind′m′;andthenx=f(p,m)★
Commented by bobhans last updated on 22/Jan/21
example : x^3 =x+6 y=(x−p)(x^3 −x−6) (dy/dx) = x^3 −x+6+(x−p)(3x^2 −1) (dy/dx)= 4x^3 −4px^2 −2x+p+6 (dy/dx)=4x+24−4px^2 −2x+p+6=2x−4px^2 +30+p let (dy/dx) = mx ⇒4px^2 +(m−2)x−(p+30)=0 x=((2−m±(√((m−2)^2 +16p(p+30))))/(8p)) now { ((p=?)),((m=?)) :}
example:x3=x+6y=(x−p)(x3−x−6)dydx=x3−x+6+(x−p)(3x2−1)dydx=4x3−4px2−2x+p+6dydx=4x+24−4px2−2x+p+6=2x−4px2+30+pletdydx=mx⇒4px2+(m−2)x−(p+30)=0x=2−m±(m−2)2+16p(p+30)8pnow{p=?m=?
Commented by ajfour last updated on 23/Jan/21
Answered by ajfour last updated on 23/Jan/21
x=−(((3c+p)(2p−3c)+3cp(m−2))/(9cp^2 +(3c+p)(m−2))) (m−2)x^2 +(2p−3c)x+(3cp)=0 say m−2=M x=−((3cpM+(3c+p)(2p−3c))/((3c+p)M+9cp^2 )) let (M/p)=s , (1/p)=q x=−((3cs+(3cq+1)(2−3cq))/((3cq+1)s+9c)) s[3cs+(3cq+1)(2−3cq)]^2 −(2−3cq)[3cs+(3cq+1)(2−3cq)][(3cq+1)s+9c] +3c[(3cq+1)s+9c]^2 =0 9c^2 s^3 +6c(3cq+1)(2−3cq)s^2 +(3cq+1)^2 (2−3cq)^2 s_(−) −6c(3cq+1)s^2 −54c^2 s +9c^2 q(3cq+1)s^2 +81c^3 qs −(3cq+1)^2 (2−3cq)^2 s_(−) −9c(3cq+1)(2−3cq)^2 +3c(3cq+1)^2 s^2 +54c(3cq+1)s +243c^3 =0 ⇒ (9c^2 )s^3 +[6c(3cq+1)(2−3cq) −6c(3cq+1)+9c^2 q(3cq+1) +3c(3cq+1)^2 ]s^2 + [54c^2 +81c^3 q+54c(3cq+1)]s +[−9c(3cq+1)(2−3cq)^2 +243c^3 ]=0 let q=0 ⇒ s^3 +(1/c)s^2 +6((1/c)+1)s −((4/c)−27c)=0 s=−2.54409 , −0.95044 for c=((30)/(19(√(19)))) x=−((3cs+(3cq+1)(2−3cq))/((3cq+1)s+9c)) and for q=0 x=−((3cs+2)/(s+9c)) = right answer is x=1.14708
x=−(3c+p)(2p−3c)+3cp(m−2)9cp2+(3c+p)(m−2)(m−2)x2+(2p−3c)x+(3cp)=0saym−2=Mx=−3cpM+(3c+p)(2p−3c)(3c+p)M+9cp2letMp=s,1p=qx=−3cs+(3cq+1)(2−3cq)(3cq+1)s+9cs[3cs+(3cq+1)(2−3cq)]2−(2−3cq)[3cs+(3cq+1)(2−3cq)][(3cq+1)s+9c]+3c[(3cq+1)s+9c]2=09c2s3+6c(3cq+1)(2−3cq)s2+(3cq+1)2(2−3cq)2s−−6c(3cq+1)s2−54c2s+9c2q(3cq+1)s2+81c3qs−(3cq+1)2(2−3cq)2s−−9c(3cq+1)(2−3cq)2+3c(3cq+1)2s2+54c(3cq+1)s+243c3=0⇒(9c2)s3+[6c(3cq+1)(2−3cq)−6c(3cq+1)+9c2q(3cq+1)+3c(3cq+1)2]s2+[54c2+81c3q+54c(3cq+1)]s+[−9c(3cq+1)(2−3cq)2+243c3]=0letq=0⇒s3+1cs2+6(1c+1)s−(4c−27c)=0s=−2.54409,−0.95044forc=301919x=−3cs+(3cq+1)(2−3cq)(3cq+1)s+9candforq=0x=−3cs+2s+9c=rightanswerisx=1.14708

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *