(d2y)/(dx2) must be integrated twice to recover y. Each integration introduces a new constant. Differentiating a constant gives zero. A linear term disappears after two derivatives, so the second derivative alone loses both its coefficient and the vertical offset. Integrate once to obtain the gradient plus A. Use the given gradient to find A. Integrate the resulting gradient and add B. Use the point on the curve to find B. y =2, y (0)=3, y(0)=4 y =2x+3 y=x2+3x+4. Exam check: one point condition does not usually determine both constants. A tangent equation supplies both its gradient and a point when the contact abscissa is known.

\(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\) must be integrated twice to recover \(y\). Each integration introduces a new constant.

Differentiating a constant gives zero. A linear term disappears after two derivatives, so the second derivative alone loses both its coefficient and the vertical offset.

Integrate once to obtain the gradient plus \(A\). Use the given gradient to find \(A\). Integrate the resulting gradient and add \(B\). Use the point on the curve to find \(B\).

\(y^{\prime\prime}=2,\ y^{\prime}(0)=3,\ y(0)=4\quad\Rightarrow\quad y^{\prime}=2x+3\quad\Rightarrow\quad y=x^2+3x+4\).

Exam check: one point condition does not usually determine both constants. A tangent equation supplies both its gradient and a point when the contact abscissa is known.

Recovering a curve from its second derivative

Recovering a curve from its second derivative