Extraordinary Differential Equations #6

Calculus Level 3

The two functions y = y ( x ) y=y(x) and w = w ( x ) w=w(x) are such that the 2 differential equations a y ′ ′ + b y ′ + c y = x 2 ay''+by'+cy=x^2 and w ′ ′ ′ + d w ′ ′ + e w ′ + f w = x 2 w'''+dw''+ew'+fw=x^2 share the same particular solution y p ( x ) = w p ( x ) = − x 2 6 + 5 x 18 − 37 108 y_p(x)=w_p(x)=-\frac{x^2}{6}+\frac{5x}{18}-\frac{37}{108} where a , b , c , d , e a, b, c, d, e and f f are integer constants to be determined.

The general solutions for both differential equations would be in the form y ( x ) = c 1 e J x + c 2 e K x − x 2 6 + 5 x 18 − 37 108 y(x)=c_1 e^{Jx}+c_2 e^{Kx}-\frac{x^2}{6}+\frac{5x}{18}-\frac{37}{108} and w ( x ) = c 3 e L x + c 4 e M x + c 5 e N x − x 2 6 + 5 x 18 − 37 108 w(x)=c_3 e^{Lx}+c_4 e^{Mx}+c_5 e^{Nx}-\frac{x^2}{6}+\frac{5x}{18}-\frac{37}{108} where c 1 c_1 through c 5 c_5 are arbitrary constants of integration, and J , K , L , M J, K, L, M and N N are real numbers to be determined.

Determine the sum 1 J + K + L + M + N \frac{1}{J+K+L+M+N} .

(This problem is part of the set Extraordinary Differential Equations .)

6 2 3 5 4 8 1 7

Wee Xian Bin by Wee Xian Bin , 25, Singapore

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1 solution

Wee Xian Bin
Jan 31, 2017

First obtain the derivatives of the particular solution y p ′ = w p ′ = − x 3 + 5 18 y_p'=w_p'=-\frac{x}{3}+\frac{5}{18} , y p ′ ′ = w p ′ ′ = − 1 3 y_p''=w_p''=-\frac{1}{3} and w p ′ ′ ′ = 0 w_p'''=0 . Thus notice that a = d , b = e , c = f a=d, b=e, c=f . Then we have: − a 3 − b x 3 + 5 b 18 − c x 2 6 + 5 c x 18 − 37 c 108 = x 2 -\frac{a}{3}-\frac{bx}{3}+\frac{5b}{18}-\frac{cx^2}{6}+\frac{5cx}{18}-\frac{37c}{108}=x^2 If we compare the coefficients separately we get the following system of linear equations:

− a 3 + 5 b 18 − 37 c 108 = 0 -\frac{a}{3}+\frac{5b}{18}-\frac{37c}{108}=0

− b 3 + 5 c 18 = 0 -\frac{b}{3}+\frac{5c}{18}=0

− c 6 = 1 -\frac{c}{6}=1

and solve to get a = 2 , b = − 5 , c = − 6 a=2, b=-5, c=-6 .

The auxiliary equation for the 1st ODE is 2 λ 2 − 5 λ − 6 = 0 ⟹ λ = 5 4 ± 73 4 2\lambda^2-5\lambda-6=0 \implies \lambda=\frac{5}{4} \pm \frac{\sqrt{73}}{4} .

The auxiliary equation for the 2nd ODE is λ 3 + 2 λ 2 − 5 λ − 6 = 0 ⟹ λ = 2 , − 1 , − 3 \lambda^3+2\lambda^2-5\lambda-6=0 \implies \lambda=2, -1, -3 .

Therefore 1 J + K + L + M + N = 1 5 4 + 5 4 + 2 − 1 − 3 = 2 \frac{1}{J+K+L+M+N}=\frac{1}{\frac{5}{4}+\frac{5}{4}+2-1-3}=2 .

1 Helpful 0 Interesting 0 Brilliant 0 Confused

in the last steps the use of vietas formula would have reduced solving time, but same way otherwise(+1)

Aareyan Manzoor - 4 years ago

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