The aim of this question is to introduce the **application of differential equations**.

Any equation that **contains one or more derivative terms** is called a **differential equation**. The solution of such an equation is not that simple, however it’s **very similar to the algebraic solution** of equations.

To solve such an equation we **first replace the derivative term** with a variable $ D $ which reduces the **differential equation to a simple algebraic equation**. Then we **solve this equation** for the **algebraic roots**. Once we have these roots, we simply use the general form of the solution to **retrieve the final solution**.

An** alternate approach** is to use the **standard textbook integration tables**. This process is further explained in the solution given below.

## Expert Answer

Let $ y $ be the required function. Then **under the given constraint:**

\[ \text{ function’s square plus the square of its derivative } = \ 1 \]

\[ \Rightarrow y^{ 2 } \ + \ \bigg ( \dfrac{ dy }{ dx } \bigg )^{ 2 } \ = \ 1 \]

**Rearranging:**

\[ \bigg ( \dfrac{ dy }{ dx } \bigg )^{ 2 } \ = \ 1 \ – \ y^{ 2 } \]

\[ \Rightarrow \dfrac{ dy }{ dx }\ = \ \pm \sqrt{ 1 \ – \ y^{ 2 } } \]

**Rearranging:**

\[ \dfrac{ 1 }{ \pm \sqrt{ 1 \ – \ y^{ 2 } } } \ dy \ = \ dx \]

**Integrating both sides:**

\[ \int \dfrac{ 1 }{ \pm \sqrt{ 1 \ – \ y^{ 2 } } } \ dy \ = \ \int dx \]

\[ \Rightarrow \pm \int \dfrac{ 1 }{ \sqrt{ 1 \ – \ y^{ 2 } } } \ dy \ = \ \int dx \]

**From integration tables:**

\[ \int \dfrac{ 1 }{ \sqrt{ 1 \ – \ y^{ 2 } } } \ dy \ = \ sin^{ -1 } y \ + \ c \]

**And:**

\[ \int dx \ = \ x \ + \ c \]

**The above equation becomes:**

\[ \pm sin^{ -1 } y \ = \ x \ + \ c \]

\[ \Rightarrow y \ = \ \pm sin( x \ + \ c ) \]

## Numerical Result

\[ y \ = \ \pm sin( x \ + \ c ) \]

## Example

If the the** square of the derivative** of a function **equals** its **square plus 1**, find the function.

Let $ y $ be the required function, then **under the given constraint:**

\[ \bigg ( \dfrac{ dy }{ dx } \bigg )^{ 2 } \ = \ y^{ 2 } \ + \ 1 \]

\[ \Rightarrow \dfrac{ dy }{ dx }\ = \ \pm \sqrt{ 1 \ + \ y^{ 2 } } \]

**Rearranging:**

\[ \dfrac{ 1 }{ \pm \sqrt{ 1 \ + \ y^{ 2 } } } \ dy \ = \ dx \]

**Integrating both sides:**

\[ \int \dfrac{ 1 }{ \pm \sqrt{ 1 \ = \ y^{ 2 } } } \ dy \ = \ \int dx \]

\[ \Rightarrow \pm \int \dfrac{ 1 }{ \sqrt{ 1 \ + \ y^{ 2 } } } \ dy \ = \ \int dx \]

**From integration tables:**

\[ \int \dfrac{ 1 }{ \sqrt{ 1 \ + \ y^{ 2 } } } \ dy \ = \ tan^{ -1 } y \ + \ c \]

**And:**

\[ \int dx \ = \ x \ + \ c \]

**The above equation becomes:**

\[ \pm tan^{ -1 } y \ = \ x \ + \ c \]

\[ \Rightarrow y \ = \ \pm tan( x \ + \ c ) \]