Question

Form differential equations to the curve ${y^2} = m\left( {{n^2} - {x^2}} \right)$ where $m$ and $n$ are arbitrary constants.

Answer 1

In the given problem, we are required to form differential equation to the curve ${y^2} = m\left( {{n^2} - {x^2}} \right)$ where m and n are some arbitrary constants. Now, we can see that there are two arbitrary constants in the given equation. So, we will aim at removing these two constants while forming the differential equation of the given equation. We will isolate both the constants one by one and then differentiate both sides of the equation in order to remove these arbitrary constants.

Complete step by step answer:
We have, ${y^2} = m\left( {{n^2} - {x^2}} \right)$
Opening the bracket, we get,
$\Rightarrow {y^2} = m{n^2} - m{x^2}$
Now, differentiating all the terms with respect to x, we get,
$\Rightarrow 2y\dfrac{{dy}}{{dx}} = 0 - 2mx$
We know that the derivative of a constant term is zero. Also we know the power rule of differentiation $\dfrac{{d\left( {{x^n}} \right)}}{{dx}} = n{x^{n - 1}}$. So, we get,
$\Rightarrow 2y\dfrac{{dy}}{{dx}} = - 2mx$
Dividing both sides of the equation by $2$, we get,
$\Rightarrow y\dfrac{{dy}}{{dx}} = - mx$
Now, shifting all the terms to left side of the equation and isolating the constant m, we get,
$\Rightarrow \dfrac{{ydy}}{{xdx}} = - m$

Now, differentiating both sides of the equation with respect to x, we get,
$\Rightarrow \dfrac{d}{{dx}}\left[ {\dfrac{y}{x}\dfrac{{dy}}{{dx}}} \right] = 0$
Again, the derivative of the constant term is zero. Also, using the quotient rule of differentiation $\dfrac{{q\left( x \right)p'\left( x \right) - p\left( x \right)q'\left( x \right)}}{{{q^2}\left( x \right)}}$, we get,
$\Rightarrow \dfrac{{x\left( {\dfrac{d}{{dx}}\left[ {y\dfrac{{dy}}{{dx}}} \right]} \right) - y\dfrac{{dy}}{{dx}}\dfrac{{d\left[ x \right]}}{{dx}}}}{{{x^2}}} = 0$
Now, cross multiplying the terms, we get,
$\Rightarrow x\left( {\dfrac{d}{{dx}}\left[ {y\dfrac{{dy}}{{dx}}} \right]} \right) - y\dfrac{{dy}}{{dx}}\dfrac{{d\left[ x \right]}}{{dx}} = 0$

Now, we know that the derivative of x with respect to x is one.
$\Rightarrow x\left( {\dfrac{d}{{dx}}\left[ {y\dfrac{{dy}}{{dx}}} \right]} \right) - y\dfrac{{dy}}{{dx}}\left( 1 \right) = 0$
Using the product rule of differentiation $\dfrac{d}{{dx}}\left[ {f\left( x \right) \times g\left( x \right)} \right] = f\left( x \right)\dfrac{d}{{dx}}\left[ {g\left( x \right)} \right] + g\left( x \right)\dfrac{d}{{dx}}\left[ {f\left( x \right)} \right]$, we get,
$\Rightarrow x\left( {y\dfrac{{{d^2}y}}{{d{x^2}}} + {{\left( {\dfrac{{dy}}{{dx}}} \right)}^2}} \right) - y\dfrac{{dy}}{{dx}}\left( 1 \right) = 0$
Opening the bracket, we get,
$\Rightarrow xy\dfrac{{{d^2}y}}{{d{x^2}}} + x{\left( {\dfrac{{dy}}{{dx}}} \right)^2} - y\dfrac{{dy}}{{dx}}\left( 1 \right) = 0$
$\therefore xy\dfrac{{{d^2}y}}{{d{x^2}}} + x{\left( {\dfrac{{dy}}{{dx}}} \right)^2} - y\dfrac{{dy}}{{dx}} = 0$

So, the differential equation to the curve ${y^2} = m\left( {{n^2} - {x^2}} \right)$ is $xy\dfrac{{{d^2}y}}{{d{x^2}}} + x{\left( {\dfrac{{dy}}{{dx}}} \right)^2} - y\dfrac{{dy}}{{dx}} = 0$.

Note: The given problem requires basic knowledge of differential equations and algebraic simplification. We also must know the derivatives of basic trigonometric functions in order to find derivatives of complex composite functions using product and quotient rules of differentiation. We also must keep in mind that we have to remove the arbitrary constants while forming the differential equation. We also know that the order of the differential equation is equal to the number of arbitrary constants in the equation. So, the order of the differential equation formed by the curve is two.