Respuesta :
Answer: D and G.
Step-by-step explanation:
For options D and G we will show that both differential equations are satisfied. For the other options we will show the pairs don't solve one of the equations.
- A. [tex] y_1 '= \cos(x)-\sin x[/tex] and [tex] y_1-2y_2= \sin x+\cos x -2(\cosx -\sin x )=3\sin x- \cos x \neq \cos x-\sin x [/tex] (when x=0 the left side is -1 and the right side is 1) so the equation [tex] y_1'=y_1 - 2y_2 [/tex] is not satisfied.
- B. [tex] y_2 '= -\sin x[/tex] and [tex] 3y_1-4y_2= 3\sin x-4\cos x \neq -\sin x[/tex] so the equation [tex] y_2'=3y_1-4y_2 [/tex] is not satisfied.
- C. [tex] y_1 '= -\sin(x)[/tex] and [tex] y_1-2y_2= \cos x -2\sin x \neq -\sin x[/tex] so these pairs don't solve the equation [tex] y_1'=y_1-2y_2 [/tex].
- D. Since [tex]y_1=y_2=e^{-x}[/tex] then [tex]y_1'=y_2'=-e^{-x}[/tex]. The first equation is satisfied, because [tex] y_1-2y_2=e^{-x}-2e^{-x}=-e^{-x}=y_1'[/tex]. The second equation is also satisfied: [tex] 3y_1-4y_2=3e^{-x}-4e^{-x}=-e^{-x}=y_2'[/tex].
- E. [tex] y_2'=e^x[/tex] and [tex] 3y_1-4y_2= 3e^x-4e^x=-e^x\neq -e^x[/tex] so they don't satisfy the equation [tex] y_2'=3y_1-4y_2 [/tex].
- F. [tex] y_1 '= 4e^{4x}[/tex] and [tex] y_1-2y_2= e^{4x}-2e^{4x}=-e^{4x} \neq 4e^{4x}[/tex], then the equation [tex] y_1'=y_1-2y_2 [/tex] is not satisfied.
- G. In this case, [tex]y_1=2e^{-2x}[/tex] and [tex]y_2=3e^{-2x}[/tex]. Computing derivatives, [tex]y_1'=-4e^{-2x}[/tex] and [tex]y_2'=-6e^{-2x}[/tex]. The first equation is satisfied, because [tex] y_1-2y_2=2e^{-2x}-6e^{-2x} =-4e^{-2x}=y_1'[/tex]. The second equation is also satisfied: [tex] 3y_1-4y_2= 6e^{-2x}-12e^{-2x}=-6e^{-2x}=y_2'[/tex].