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Consider the initial value problem $y' + 5y = 5t, y(0) = y_0$. Find the value of $y_0$ for which the solution touches, but does not cross, the $t$-axis.

This is linear first order DE, so I found integrating factor=$e^{(2/3)t}$ Solving ODE I got, $y=\frac{21}{8}-\frac{3}{4}t+Ce^{(-2/3)t}$ Using, $y(0)=y_0$, we get $C=y_0-\frac{21}{8}$. Now, if we take first derivative of the solution function $y$ to find the point where it has zero slope, we end up with two unknown in the equation, $y_0$ and $t$. $y'=\frac{-3}{4}+\frac{-2}{3}(y_0-\frac{21}{8})e^{-(2/3)t}=0.$

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