Question

Suppose that ${\mathit{y}}_{\mathbf{1}}\mathbf{,}{\mathit{y}}_{\mathbf{2}}\mathbf{,}\mathbf{\dots }\mathbf{,}{\mathit{y}}_{\mathbf{k}}$arenontrivial solutions of a homogeneous linear nth-order differential equation with constant coefficients and that$\mathbf{k}\mathbf{=}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{.}$Is the set of solutions ${\mathit{y}}_{\mathbf{1}}\mathbf{,}{\mathit{y}}_{\mathbf{2}}\mathbf{,}\mathbf{\dots }\mathbf{,}{\mathit{y}}_{\mathbf{k}}$linearly dependent or linearly independent on$\mathbf{(}\mathbf{-}\mathbf{\infty }\mathbf{,}\mathbf{\infty }\mathbf{)}$? Discuss.

Short answer

The set of solutions $y_1,y_2,\dots ,y_k$is linearly dependent on $(-\infty ,\infty )$$(-\infty ,\infty )$.

Step-by-step solution

Superposition principle-Homogeneous equations

• In the next theorem we see that the sum, or superposition, of two or more solutions of a homogeneous linear differential equation is also a solution.
• Let$y_1,y_2,\dots ,y_k$ be solutions of the homogeneous n-th-order differential equation on an interval I. Then the linear combination $y=c_1{y}_1\left(x\right)+c_2{y}_2\left(x\right)+\cdots +c_k{y}_k\left(x\right),$Where, the$c_i,i=1,2,\dots ,k$ are arbitrary constants, is also a solution on the interval.

Find the solution is linearly dependent or independent.

• A homogeneous linear differential equation of$n^{\text{th}}$order has the fundamental set of solutions having exactly n independent non-trivial (non-zero) solutions.
• All other solutions linear multiple or the linear combination of the solutions in the fundamental set and are hence dependent solutions.
• If a linear homogeneous differential equation of $n^{\text{th}}$order has n + 1 non-trivial solutions then the n + 1 solutions (as in the question) must be a dependent list of solutions because the equation can only have an independent set of solutions.
• There will be at least one solution in the n + 1 solutions that will be a linear multiple or linear combination of the other solution/s in the set.

Therefore, the set of solutions $y_1,y_2,\dots ,y_k$is linearly dependent $(-\infty ,\infty )$.