Question Number 1023 by 123456 last updated on 20/May/15

φ_n :R→R  n∈N^∗   φ_n =t^n (d^n φ/dt^n )  φ_1 (t)=?,φ_1 (1)=+1  φ_2 (t)=?,φ_2 (1)=−1,φ_2 ^′ (1)=+1  φ_3 (t)=?,φ_3 (1)=+1,φ_3 ^′ (1)=−1,φ_3 ^(′′) (1)=+1

Commented byprakash jain last updated on 21/May/15

φ_1 (t)=t((dφ_1 (t))/dt)⇒ln y=ln t+C  φ_1 (t)=kt  φ_1 (1)=1⇒k=1  φ_1 (t)=t

Answered by prakash jain last updated on 21/May/15

φ_2 (t)=t^2 (d^2 φ_2 /dt^2 )  t=e^x   (dφ_2 /dt)=e^(−x) (dφ_2 /dx), (d^2 φ_2 /dt)=((d^2 φ_2 /dx^2 )−(dφ_2 /dx))e^(−2x)   φ_2 =φ_2 ^(′′) −φ_2 ^′   φ_2 ^(′′) −φ_2 ^′ −φ_2 =0  characteristic equation  r^2 −r−1=0  r_1 ,r_2 =((1±(√5))/2)  φ_2 (x)=c_1 e^(r_1 x) +c_2 e^(r_2 x)   φ_2 (t)=c_1 t^r_1  +c_2 t^r_2    φ_2 (t)=−1⇒c_1 +c_2 =−1  φ_2 ′=c_1 r_1 t^(r_1 −1) +c_2 r_2 t^(r_2 −1)   φ_2 ′(1)=1  1=c_1 r_1 +c_2 r_2 =c_1 (((1+(√5))/2))−c_1 (((1−(√5))/2))−((1−(√5))/2)  1+((1−(√5))/2)=c_1 (√5)⇒c_1 =((3−(√5))/(√5))  c_2 =−1−c_1 =−1−((3−(√5))/(√5))=((2(√5)−3)/2)  φ_2 (t)=c_1 x^r_1  +c_2 x^r_2  , r_1 =((1+(√5))/2), r_2 =((1−(√5))/2)