Question on a differential equationLet&nbsp;x:R→R&nbsp;a solution of the differenzial equation:5x&nbsp;″(t)+10x′(t)+6x(t)=0Proof that the function:f:ℝ→ℝ

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Question on a differential equationLet&amp;nbsp;x:R→R&amp;nbsp;a solution of the differenzial equation:5x&amp;nbsp;″(t)+10x′(t)+6x(t)=0Proof that the function:f:ℝ→ℝ

Alisha Chambers · Accepted Answer

Step 1Let y:R→R be defined as follows:y(t)=t21+t4Consider that x exists; x(t)=e−t(Acos⁡(t5)+Bcos⁡(t5)) fulfills the differential equation. Consider also thatf(t)=y(x(t))Now calculate the first and second derivatives of y:y′(t)=2t(1+t4)−t2(4t3)(1+t4)2y′(t)=2t(1−t4)(1+t4)2=2t(1−t2)(1+t2)(1+t4)2andy″(t)=(2−10t4)(1+2t4+t8)−(2t−2t5)(8t7+8t3)(1+t4)4y″(t)=2+4t4+2t8−10t4−20t8−10t12+16t12+16t8−16t8−16t4(1+t4)4y″(t)=2−22t4−18t8+6t12(1+t4)4y″(t)=2(1+t4)(1−12t4+3t8)(1+t4)4y″(t)=2(1−12t4+3t8)(1+t4)3Consider that the stationary points of y (points where y′(t)=0) are −1,0,1 and of those points, only -1 and 1 yield negative values for the second derivative. Therefore the points making y maximal are -1 and 1.Now it only remains to prove that x(t) reaches the value -1 or 1 somewhere on R, which should be fairly easy (with the intermediate value theorem). After that is proven, you have a maximal value of f, which is 12.Exception: if A=B=0, then x(t) does not reach -1 or 1, but f becomes constant and thus has a maximum.

Question on a differential equation Let \(\displaystyle{x}:{\mathbb{{{R}}}}\rightarrow{\mathbb{{{R}}}}\) a

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