89 Proof

By repeatedly applying Proposition 73, we have

$$ℒ(y^{\prime })(s)=sℒ(y)(s)-y(0)$$

$$ℒ(y^{\prime \prime })(s)=sℒ(y^{\prime })(s)-y^{\prime }(0)$$

$$ℒ(y^{(n)})(s)=sℒ(y^{(n-1)})(s)-y^{(n-1)}(0),$$

so we can substitute backwards and get

$$ℒ(y^{\prime \prime })(s)=s^2ℒ(y)(s)-sy(0)-y^{\prime }(0)$$

$$ℒ(y^{\prime \prime \prime })(s)=s^3ℒ(y)(s)-s^{2}y(0)-s{y}^{\prime }(0)-y^{\prime \prime }(0)$$

and thus obtain $q_{n-1}(s)$ with coefficients $p_j=y^{(j)}(0)$ as in the initial conditions $y(0),\mathrm{\dots },y^{(n-1)}(0)$.