Find a function whose square plus the square of its derivative is 1.

Find A Function Whose Square Plus The Square Of Its Derivative Is 1

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:

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

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

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