The two independent real solutions corresponding to the complex roots m = -1 ± i are:

Boost your skills in A Level Further Mathematics Core Pure. Study with flashcards and multiple choice questions, each question is followed by hints and explanations. Get prepared for your exam with confidence!

Multiple Choice

The two independent real solutions corresponding to the complex roots m = -1 ± i are:

When an Euler-Cauchy equation yields complex roots m = α ± iβ, the real solutions come from combining those roots as y = x^{α} cos(β ln x) and y = x^{α} sin(β ln x). This works because x^{m} = e^{m ln x} = x^{α} e^{± iβ ln x} = x^{α}[cos(β ln x) ± i sin(β ln x)], so the real and imaginary parts give two independent real solutions. Here α = -1 and β = 1, so the two independent real solutions are x^{-1} cos(ln x) and x^{-1} sin(ln x). The other forms would correspond to different α or to hyperbolic functions, which don’t match the given roots.

When an Euler-Cauchy equation yields complex roots m = α ± iβ, the real solutions come from combining those roots as y = x^{α} cos(β ln x) and y = x^{α} sin(β ln x). This works because x^{m} = e^{m ln x} = x^{α} e^{± iβ ln x} = x^{α}[cos(β ln x) ± i sin(β ln x)], so the real and imaginary parts give two independent real solutions.

Here α = -1 and β = 1, so the two independent real solutions are x^{-1} cos(ln x) and x^{-1} sin(ln x). The other forms would correspond to different α or to hyperbolic functions, which don’t match the given roots.

More practice questions