diff --git a/docs/en/mathematics/ordinary-differential-equations/laplace-transform.md b/docs/en/mathematics/ordinary-differential-equations/laplace-transform.md
index d671a94..e323f1d 100644
--- a/docs/en/mathematics/ordinary-differential-equations/laplace-transform.md
+++ b/docs/en/mathematics/ordinary-differential-equations/laplace-transform.md
@@ -99,13 +99,22 @@ $$
     For large enough $s$, the case $n=1$ follows by integration by parts
 
     $$
-    \begin{align*} \mathcal{L}[f'](s) &= \int_0^\infty e^{-st} f'(t)dt, \\ &= \Big[e^{-st} f(t) \Big]_0^\infty + s\int_0^\infty e^{-st}f(t), \\ &= sF(s) - f(0) \end{align*},
+    \begin{align*} 
+        \mathcal{L}[f'](s) &= \int_0^\infty e^{-st} f'(t)dt, \\ 
+        &= \Big[e^{-st} f(t) \Big]_0^\infty + s\int_0^\infty e^{-st}f(t), \\ 
+        &= sF(s) - f(0),
+    \end{align*}
     $$
 
-    suppose $\mathcal{L}[f^{k}](s) = s^k F(s) - \sum_{r=0}^{k-1} s^r f^{(k-1-r)}(0)$ is true for $k \in \mathbb{N}$, then by assumption
+    suppose $\mathcal{L}[f^{k}](s) = s^k F(s) - \sum_{r=0}^{k-1} s^r f^{(k-1-r)}(0)$ is true for $k \in \mathbb{N}$. Then by assumption
 
     $$
-    \begin{align*} \mathcal{L}[f^{k+1}](s) &= \int_0^\infty e^{-st} f^{(k+1)}(t)dt , \\ &= \Big[e^{-st} f^{(k+1)}(t) \Big]_0^\infty + s\int_0^\infty e^{-st}f^{(k)}(t), \\ &= s \mathcal{L}[f^{(k)}] - f^{(k)}(0), \\ &= s \Big(s^k F(s) - \sum_{r=0}^{k-1} s^r f^{(k-1-r)}(0)\Big) - f^{(k)}(0), \\ &= s^{k+1} F(s) - \sum_{r=0}^{k} s^r f^{(k-r)}(0) \end{align*}.
+    \begin{align*} 
+        \mathcal{L}[f^{k+1}](s) &= \int_0^\infty e^{-st} f^{(k+1)}(t)dt, \\ 
+        &= \Big[e^{-st} f^{(k+1)}(t) \Big]_0^\infty + s\int_0^\infty e^{-st}f^{(k)}(t), \\ 
+        &= s \mathcal{L}[f^{(k)}] - f^{(k)}(0), \\ &= s \Big(s^k F(s) - \sum_{r=0}^{k-1} s^r f^{(k-1-r)}(0)\Big) - f^{(k)}(0), \\ 
+        &= s^{k+1} F(s) - \sum_{r=0}^{k} s^r f^{(k-r)}(0).
+    \end{align*}
     $$
 
 ## Examples