Verifying Solutions, Second Derivatives and Tangent-Line Estimates

Verifying a solution

A function is a solution of a differential equation when substituting it and its derivatives makes the equation an identity for every $x$ in the domain. To verify that $y=3e^{2x}-x-\frac12$ solves $\dfrac{dy}{dx}=2y+2x$: left side $y'=6e^{2x}-1$; right side $2\left(3e^{2x}-x-\frac12\right)+2x=6e^{2x}-1$. The two sides agree identically, so it is a solution. Checking at a single point is not a verification, and neither is solving the equation from scratch.

The second derivative from the equation

Given $\dfrac{dy}{dx}=F(x,y)$, differentiate both sides with respect to $x$, remembering that every $y$ carries a factor $\dfrac{dy}{dx}$, then replace $\dfrac{dy}{dx}$ by $F(x,y)$ so that the result contains only $x$ and $y$. Example: $\dfrac{dy}{dx}=x-y$ gives $\dfrac{d^2y}{dx^2}=1-\dfrac{dy}{dx}=1-(x-y)=1-x+y$.

Tangent-line estimate and its direction

The equation hands you the slope at the initial point. With $y(1)=2$ and $\dfrac{dy}{dx}=x-y$: slope $1-2=-1$, tangent line $y=2-(x-1)$, so $y(1.2)\approx1.8$. Over- or underestimate comes from $\dfrac{d^2y}{dx^2}$ at the point: $1-1+2=2\gt0$, so the solution curve is concave up there, the tangent line lies below the curve, and $1.8$ is an underestimate. The rubric wants all three: the sign of $\dfrac{d^2y}{dx^2}$, the word "concave", and the conclusion.