The aim of this question is to learn how to **solve** simple ordinary **differential equations** and then use them to solve different **word problems**.

A **differential equation** is an equation that **involves derivatives** and requires **integration** during their solution.

While solving such equations, we may encounter **integration constants** that are calculated using the **initial conditions** given in the question.

## Expert Anwer

**Given:**

\[ \dfrac{ dL }{ dx } \ = \ -kL \]

**Rearranging:**

\[ \dfrac{ 1}{ L } \ dL \ = \ -k \ dx \]

**Integrating both sides:**

\[ \int \ \dfrac{ 1}{ L } \ dL \ = \ -k \ \int \ dx \]

**Using integration tables:**

\[ \int \ \dfrac{ 1}{ L } \ dL \ = \ ln| \ L \ | \ \text{ and } \ \int \ dx \ = \ x \]

Substituting these values in the above equation:

\[ ln| \ L \ | \ = \ -k \ x \ … \ … \ … \ (1) \]

**Exponentiating both sides:**

\[ e^{ ln| \ L \ | } \ = \ e^{ -k \ x } \]

Since:

\[ e^{ ln| \ L \ | } \ = \ L \]

So, the above equation becomes:

\[ L \ = \ e^{ -k \ x } \ … \ … \ … \ (2) \]

Given the following **initial condition**:

\[ L \ = \ 0.5 \ at \ x \ = \ 18 \ ft \]

Equation (1) becomes:

\[ ln| \ 0.5 \ | \ = \ -k \ ( \ 18 \ ) \]

\[ \Rightarrow k = \dfrac{ ln| \ 0.5 \ | }{ -18 } \]

\[ \Rightarrow k = 0.0385 \]

Substitute this value in the equation (1) and (2):

\[ ln| \ L \ | \ = \ -0.0385 \ x \ … \ … \ … \ (3) \]

And:

\[ L \ = \ e^{ -0.0385 \ x } \ … \ … \ … \ (4) \]

To find the depth $x$ at which intensity $L$ falls to **one-tenth**, we put the following values in the equation (3):

\[ ln| \ 0.1 \ | \ = \ -0.0385 \ x \]

\[ \Rightarrow x \ = \ \dfrac{ ln| \ 0.1 \ | }{ -0.0385 } \]

\[ \Rightarrow x \ = \ 59.8 \ ft \]

## Numerical Result

\[ x \ = \ 59.8 \ ft \]

## Example

In the above question, with the **same differential equation and initial condition**, find the **depth at which intensity reduces** to 25% and 75%.

**Part(a):** Substitute $ L = 0.25 $ in equation no. (3):

\[ ln| \ 0.25 \ | \ = \ -0.0385 \ x \]

\[ \Rightarrow x \ = \ \dfrac{ ln| \ 0.25 \ | }{ -0.0385 } \]

\[ \Rightarrow x \ = \ 36 \ ft \]

**Part(b):** Substitute $ L = 0.75 $ in equation no. (3):

\[ ln| \ 0.75 \ | \ = \ -0.0385 \ x \]

\[ \Rightarrow x \ = \ \dfrac{ ln| \ 0.75 \ | }{ -0.0385 } \]

\[ \Rightarrow x \ = \ 7.47 \ ft \]