99. Prove that $x^{2} - y^{2} = c {(x^{2} + y^{2})}^{2}$ is the general equation of the differential equation $(x^{3} - 3 x y^{2}) d x = (y^{3} - 3 x^{2} y) d y$ where c is a parameter.

19 Views|Posted 8 months ago

Asked by Shiksha User

1 Answer

Sort by:

Answered by

Vishal Baghel | Contributor-Level 10

8 months ago

$\Rightarrow \frac{d y}{d x} = \frac{x^{3} - 3 x y^{2}}{y^{3} - 3 x^{2} y} . . . . . . . . . . (1)$

This is a homogenous equation. To simplify it, we need to make the substitution as:

$\begin{array}{l} y = v x \\ \Rightarrow \frac{d}{d x} (y) = \frac{d}{d x} (v x) \\ \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x} \end{array}$

Substituting the values of y and $\frac{d v}{d x}$ in equation (1), we get:

$\begin{array}{l} v + x \frac{d v}{d x} = \frac{x^{3} - 3 x {(v x)}^{2}}{{(v x)}^{3} - 3 x^{2} (v x)} \\ \Rightarrow v + x \frac{d v}{d x} = \frac{1 - 3 v^{2}}{v^{3} - 3 v} \\ \Rightarrow x \frac{d v}{d x} = \frac{1 - 3 v^{2}}{v^{3} - 3 v} - v \\ \Rightarrow x \frac{d v}{d x} = \frac{1 - 3 v^{2} - v (v^{3} - 3 v)}{v^{3} - 3 v} \\ \Rightarrow x \frac{d v}{d x} = \frac{1 - v^{4}}{v^{3} - 3 v} \\ \Rightarrow (\frac{v^{3} - 3 v}{1 - v^{4}}) d v = \frac{d x}{x} \end{array}$

Integrating both sides, we get:

$\begin{array}{l} \int (\frac{v^{3} - 3 v}{1 - v^{4}}) d v = l o g x + l o g C^{'} . . . . . . . . . (2) \\ N o w, \int (\frac{v^{3} - 3 v}{1 - v^{4}}) d v = \int \frac{v^{3} d v}{1 - v^{4}} - 3 \int \frac{v d v}{1 - v^{4}} \\ \Rightarrow \int (\frac{v^{3} - 3 v}{1 - v^{4}}) d v = I_{1} - 3 I_{2}, W h e r e, I_{1} = \int \frac{v^{3} d v}{1 - v^{4}} a n d I_{2} = \int \frac{v d v}{1 - v^{4}} . . . . . . . . . . . (3) \end{array}$

$\begin{array}{l} L e t, 1 - v^{4} = t . \\ ∴ \frac{d}{d v} (1 - v^{4}) = \frac{d t}{d v} \\ \Rightarrow - 4 v^{3} = \frac{d t}{d v} \\ \Rightarrow v^{3} d v = - \frac{d t}{4} \\ N o w, I_{1} = \int - \frac{d t}{4} = - l o g t = - \frac{1}{4} l o g (1 - v^{4}) \end{array}$

$\begin{array}{l} A n d, I_{2} = \int \frac{v d v}{1 - v^{4}} = \int \frac{v d v}{1 - (v^{2})^{2}} \\ L e t, v^{2} = p . \\ ∴ \frac{d}{d v} (v^{2}) = \frac{d p}{d v} \\ \Rightarrow 2 v = \frac{d p}{d v} \\ \Rightarrow v d v = \frac{p}{2} \\ \Rightarrow I_{2} = \frac{1}{2} \int \frac{d p}{1 - p^{2}} = \frac{1}{2 * 2} l o g | \frac{1 + p}{1 - p} | = \frac{1}{4} l o g | \frac{1 + v^{2}}{1 - v^{2}} | \end{array}$

Substituting the values of $I_{1}$ and $I_{2}$ in equation (3), we get:

$\int (\frac{v^{3} - 3 v}{1 - v^{4}}) d v = - \frac{1}{4} l o g (1 - v^{4}) - \frac{3}{4} l o g | \frac{1 + v^{2}}{1 - v^{2}} |$

Therefore, equation (2) becomes:

$\begin{array}{l} \frac{1}{4} l o g (1 - v^{4}) - \frac{3}{4} l o g | \frac{1 + v^{2}}{1 - v^{2}} | = l o g x + l o g C^{'} \\ \Rightarrow - \frac{1}{4} l o g [(1 - v^{4}) (\frac{1 + v^{2}}{1 - v^{2}})] = l o g C^{'} x \\ \Rightarrow \frac{{(1 + v^{2})}^{4}}{{(1 - v^{2})}^{2}} = {(C^{'} x)}^{- 4} \\ \Rightarrow \frac{{(1 + \frac{y^{2}}{x^{2}})}^{4}}{{(1 - \frac{y^{2}}{x^{2}})}^{2}} = \frac{1}{C^{' 4} x^{4}} \\ \Rightarrow \frac{{(x^{2} + y^{2})}^{4}}{x^{4} {(x^{2} - y^{2})}^{2}} = \frac{1}{C^{' 4} x^{4}} \\ \Rightarrow {(x^{2} - y^{2})}^{2} = C^{' 4} {(x^{2} + y^{2})}^{4} \\ \Rightarrow (x^{2} - y^{2}) = C^{' 2} {(x^{2} + y^{2})}^{2} \\ \Rightarrow x^{2} - y^{2} = C {(x^{2} + y^{2})}^{2}, w h e r e C = C^{' 2} \end{array}$