Proving single solution to an initial value problemy'=|11+x2+sin|x2+arctany2||,&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;y(x0)=y0or each&nbsp;(x0,y0)∈R×R&nbsp;I need to prove that there is a single solution defined on&nbsp;R

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Proving single solution to an initial value problemy&#039;=|11+x2+sin|x2+arctany2||,&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;y(x0)=y0or each&amp;nbsp;(x0,y0)∈R×R&amp;nbsp;I need to prove that there is a single solution defined on&amp;nbsp;R

Zemmiq34 · Accepted Answer

Step 1Letg(x,y)=sin|x2+arctan(y2)|=sin(x2+arctan(y2))This function is derivable, hencegy(x,y)=2yy4+1cos⁡{(x2+arctan⁡{(y2)})}gy&amp;nbsp;is continuous thus for each J there is a constant&amp;nbsp;LJ&amp;nbsp;so that|g(x,y1)−g(x,y2)|≤LJ|y1−y2|therefore|f(x,y1)−f(x,y2)|=||11+x2+g(x,y1)|−|11+x2+g(x,y2)∣∣≤|11+x2+g(x,y1)−11+x2−g(x,y2)|=|g(x,y1)−g(x,y2)|≤LJ|y1−y2|and&amp;nbsp;f(x,y)&amp;nbsp;is Lipschitz continuous on the 𝑦 variable, and this is true for each box&amp;nbsp;J×(−∞,∞)&amp;nbsp;and therefore there is a single solution on&amp;nbsp;R&amp;nbsp;for each&amp;nbsp;(x0,y0)∈R×R

Proving single solution to an initial value problem y' = \left|\frac{1}{1+x^2}

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