Constant of Integration — Question 4

Question 4 - Solution

Let $y$ be any solution of the differential equation $$y'=f(x)$$ on the interval $I$.

Since $F'(x)=f(x)$ and $y'(x)=f(x)$ for all $x\in I$, we have $$y'(x)-F'(x)=0 \quad\text{for all }x\in I.$$

Consider the function $$H(x)=y(x)-F(x).$$

Differentiate $H(x)$: $$H'(x)=y'(x)-F'(x)=0 \quad\text{for all }x\in I.$$

Thus, $H'(x)=0$ on $I$.

By the result that a function with zero derivative on an interval is constant, there exists a real number $C$ such that $$H(x)=C \quad\text{for all }x\in I.$$

Substituting back gives $$y(x)-F(x)=C.$$

Rewriting, $$y=F(x)+C.$$

$$\boxed{y=F(x)+C}$$