Consider the problemY′=d1t+1+c⋅tan−1(Y(t))−d12,Y(0)=0with c a given constant. Since Y′(0)=d12, the solution Y (t) is initially increasing as t increases, regardless of the value of c. As best you can, show that there is a value of c, call it c∗, for which (1) if c&gt;c⋅, the solution Y(t) increases indefinitely, and (2) if c&lt;c⋅, then Y(t) increases initially, but then peaks and decreases. Using ode45, determine c∗ to within 0.00005, and then calculate the associated solution Y(t) for 0≤t≤50.

Question

Consider the problemY′=d1t+1+c⋅tan−1(Y(t))−d12,Y(0)=0with c a given constant. Since Y′(0)=d12, the solution Y (t) is initially increasing as t increases, regardless of the value of c. As best you can, show that there is a value of c, call it c∗, for which (1) if c&amp;gt;c⋅, the solution Y(t) increases indefinitely, and (2) if c&amp;lt;c⋅, then Y(t) increases initially, but then peaks and decreases. Using ode45, determine c∗ to within 0.00005, and then calculate the associated solution Y(t) for 0≤t≤50.

kachnaemra · Accepted Answer

Step 1c=0&amp;nbsp;can be integrated directly,y(t)=y0(t)=ln⁡(1+t)−t2,and the value at&amp;nbsp;t=e2−1&amp;gt;6&amp;nbsp;is negative.For any&amp;nbsp;c&amp;gt;0, the observed initial slope gives an initial segment where y is positive.Step 2For&amp;nbsp;C≈∞, even a small positive value of y will lead to a large slope catapulting the y curve towards very large values. After that the other two terms do not matter any more, the slope remains positive, y does not come down, back from &quot;almost infinity&quot;.One could try to make the last point a bit more quantitative by replacing &quot;almost infinity&quot; with some very large value, e.g.,&amp;nbsp;c=1000, and estimate the magnitude of all terms for&amp;nbsp;t=0.01, to then find the announced behavior.

Consider the problem \(\displaystyle{Y}'={d}{\frac{{{1}}}{{{t}+{1}}}}+{c}\cdot{{\tan}^{{-{1}}}{\left({Y}{\left({t}\right)}\right)}}-{d}{\frac{{{1}}}{{{2}}}},\quad{Y}{\left({0}\right)}={0}\) with c a given constant.

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