Showing that continuous and differentiable s(t), with&nbsp;s(0)=0&nbsp;and&nbsp;s′(t)≤2ts(t)+s(t)&nbsp;for&nbsp;t&gt;0, is identically zero for&nbsp;t≥0.Let&nbsp;s:[0,+∞)→[0,+∞)&nbsp;be a continuous function that is differentiable on&nbsp;(0,+∞).It also holds that&nbsp;s'(t)≤2ts(t)+s(t),&nbsp;&nbsp;t&gt;0&nbsp;&nbsp;&nbsp;&nbsp;s(0)=0Show that&nbsp;s(t)=0,t≥0.Do we us some theorem, maybe Rolle or the mean value theorem?Do we have to check maybe the sign of&nbsp;2ts(t)+s(t)? This depends on the value of s(t), or not?

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Showing that continuous and differentiable s(t), with&amp;nbsp;s(0)=0&amp;nbsp;and&amp;nbsp;s′(t)≤2ts(t)+s(t)&amp;nbsp;for&amp;nbsp;t&amp;gt;0, is identically zero for&amp;nbsp;t≥0.Let&amp;nbsp;s:[0,+∞)→[0,+∞)&amp;nbsp;be a continuous function that is differentiable on&amp;nbsp;(0,+∞).It also holds that&amp;nbsp;s&#039;(t)≤2ts(t)+s(t),&amp;nbsp;&amp;nbsp;t&amp;gt;0&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;s(0)=0Show that&amp;nbsp;s(t)=0,t≥0.Do we us some theorem, maybe Rolle or the mean value theorem?Do we have to check maybe the sign of&amp;nbsp;2ts(t)+s(t)? This depends on the value of s(t), or not?

Brianna Sims · Accepted Answer

Step 1Set s(t)=14exp⁡(t2)(∫0texp⁡(−τ22)dτ)2(t≥0). Then s(t)≥0,s(0)=0 ands′(t)=t2exp⁡(t2)(∫0texp⁡(−τ22)dτ)2+14exp⁡(t2)2∫0texp⁡(−τ22)dτexp⁡(−t22)Step 2=2ts(t)+12exp⁡(t22)∫0texp⁡(−τ22)dτ=2ts(t)+s(t)Edit: The idea to find this solution: Consider the ODE s′=2ts+s. Then substitute u=exp⁡(−t2)s to obtain u′=exp⁡(−t22)u. Separation of variables then leads to the maximal solution of the ivp for u hence for s.

Showing that continuous and differentiable s(t), with s(0)=0 and s'(t)\leq 2ts(t)+\sqrt{s(t

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