Q33E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Use the method in Problem 32 to find the orthogonal trajectories for each of the given families of curves, where k is a parameter.

 (a) $\mathbf{2}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{=}\mathbf{k}$

(b) $\mathbf{y}\mathbf{=}{\mathbf{kx}}^{\mathbf{4}}$

(c) $\mathbf{y}\mathbf{=}{\mathbf{e}}^{\mathbf{kx}}$

(d)  ${\mathbf{y}}^{\mathbf{2}}\mathbf{=}\mathbf{kx}$

[Hint: First express the family in the form F(x, y) = k.]

Step-by-Step Solution

Answer

1. $\mathbf{x}\mathbf{=}c{\mathbf{y}}^{\mathbf{2}}$
   
2. ${\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{4}{\mathbf{y}}^{\mathbf{2}}\mathbf{=}c$
   
3. $\mathbf{2}{\mathbf{y}}^{\mathbf{2}}\mathbf{ln} \mathbf{y}\mathbf{-}{\mathbf{y}}^{\mathbf{2}}+\mathbf{2}{\mathbf{x}}^{\mathbf{2}}=c$
   
4. $\mathbf{2}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{=}\mathbf{C}$
   

(a): Find the orthogonal trajectory of curve 2 x 2 + y 2 = k

Here, the curve is $\mathbf{2}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{=}\mathbf{k}$

$\begin{array}{l}\frac{\partial \mathbf{F}}{\partial \mathbf{y}}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{=}\mathbf{2}\mathbf{y}\\ \frac{\partial \mathbf{F}}{\partial \mathbf{x}}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{=}\mathbf{-}\mathbf{4}\mathbf{x}\\ \mathbf{2}\mathbf{y} \mathbf{dx}\mathbf{-}\mathbf{4}\mathbf{x} \mathbf{dy}\mathbf{=}\mathbf{0}\\ \int \frac{\mathbf{1}}{\mathbf{x}}\mathbf{dx}\mathbf{=}\mathbf{2}\int \frac{\mathbf{1}}{\mathbf{y}}\mathbf{dy}\mathbf{=}\mathbf{0}\\ \mathbf{ln} \left|\mathbf{x}\right|=\mathbf{2}\mathbf{ln} \left|\mathbf{y}\right|\mathbf{+}c\\ \mathbf{x}\mathbf{=}c{\mathbf{y}}^{\mathbf{2}}\end{array}$

Hence the solution is $\mathbf{x}\mathbf{=}c{\mathbf{y}}^{\mathbf{2}}$

(b): Determine the orthogonal trajectory of curve y = k x 4 .

Here the curve is $y=k{x}^4$

$\begin{array}{l}\frac{\partial \mathbf{F}}{\partial \mathbf{y}}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{x}}^{\mathbf{4}}}\\ \frac{\partial \mathbf{F}}{\partial \mathbf{x}}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{=}\mathbf{-}\frac{\mathbf{4}\mathbf{y}}{{\mathbf{x}}^{\mathbf{5}}}\\ \frac{\mathbf{1}}{{\mathbf{x}}^{\mathbf{4}}}\mathbf{dx}\mathbf{+}\frac{\mathbf{4}\mathbf{y}}{{\mathbf{x}}^{\mathbf{5}}}\mathbf{dy}\mathbf{=}\mathbf{0}\\ \int \mathbf{x} \mathbf{dx}\mathbf{=}\mathbf{-}\mathbf{4}\int \mathbf{y} \mathbf{dy}\mathbf{=}\mathbf{0}\\ \frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{2}{\mathbf{y}}^{\mathbf{2}}\mathbf{+}c\\ {\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{4}{\mathbf{y}}^{\mathbf{2}}\mathbf{=}c\end{array}$

Hence the solution is ${\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{4}{\mathbf{y}}^{\mathbf{2}}\mathbf{=}c$

(c): Evaluate the orthogonal trajectory of curve y = e kx

Here the curve is $y=k{e}^x$

$\begin{array}{l}\frac{\mathbf{lny}}{\mathbf{x}}\mathbf{=}\mathbf{k}\\ \frac{\partial \mathbf{F}}{\partial \mathbf{y}}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{xy}}\\ \frac{\partial \mathbf{F}}{\partial \mathbf{x}}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{=}\mathbf{-}\frac{\mathbf{ln} \mathbf{y}}{{\mathbf{x}}^{\mathbf{2}}}\\ \frac{\mathbf{1}}{\mathbf{xy}}\mathbf{dx}\mathbf{+}\frac{\mathbf{ln} \mathbf{y}}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{dy}\mathbf{=}\mathbf{0}\\ \int \mathbf{x} \mathbf{dx}\mathbf{=}\mathbf{-}\int \mathbf{y} \mathbf{ln} \mathbf{y} \mathbf{dy}\mathbf{=}\mathbf{0}\\ \frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{2}}\mathbf{=}\frac{\mathbf{-}{\mathbf{y}}^{\mathbf{2}} \mathbf{ln} \mathbf{y}}{\mathbf{2}}\mathbf{+}\frac{{\mathbf{y}}^{\mathbf{2}}}{\mathbf{4}}\mathbf{+}c\\ \frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{2}}\mathbf{+}\frac{{\mathbf{y}}^{\mathbf{2}} \mathbf{ln} \mathbf{y}}{\mathbf{2}}\mathbf{-}\frac{{\mathbf{y}}^{\mathbf{2}}}{\mathbf{4}}\mathbf{+}c\\ \mathbf{2}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{2}{\mathbf{y}}^{\mathbf{2}} \mathbf{ln} \mathbf{y}\mathbf{-}{\mathbf{y}}^{\mathbf{2}}\mathbf{=}c\\ \mathbf{2}{\mathbf{y}}^{\mathbf{2}} \mathbf{ln} \mathbf{y}\mathbf{-}{\mathbf{y}}^{\mathbf{2}}+\mathbf{2}{\mathbf{x}}^{\mathbf{2}}=c\end{array}$

Hence the solution is $\mathbf{2}{\mathbf{y}}^{\mathbf{2}} \mathbf{ln} \mathbf{y}\mathbf{-}{\mathbf{y}}^{\mathbf{2}}+\mathbf{2}{\mathbf{x}}^{\mathbf{2}}=c$

(d): Find the orthogonal trajectory of curve y 2 = k x

Here the curve is $y^2=kx$.

$\begin{array}{l}\frac{{\mathbf{y}}^2}{\mathbf{x}}\mathbf{=}\mathbf{k}\\ \frac{\partial \mathbf{F}}{\partial \mathbf{y}}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{=}\frac{\mathbf{2}\mathbf{y}}{\mathbf{x}}\\ \frac{\partial \mathbf{F}}{\partial \mathbf{x}}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{=}\mathbf{-}\frac{{\mathbf{y}}^{\mathbf{2}}}{{\mathbf{x}}^{\mathbf{2}}}\\ \frac{\mathbf{2}\mathbf{y}}{\mathbf{x}}\mathbf{dx}\mathbf{+}\frac{{\mathbf{y}}^{\mathbf{2}}}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{dy}\mathbf{=}\mathbf{0}\\ \mathbf{2}\int \mathbf{xdx}\mathbf{=}\mathbf{-}\int \mathbf{ydy}\mathbf{=}\mathbf{0}\\ \frac{\mathbf{2}{\mathbf{x}}^{\mathbf{2}}}{\mathbf{2}}\mathbf{=}\mathbf{-}\frac{{\mathbf{y}}^{\mathbf{2}}}{\mathbf{2}}\mathbf{+}c\\ \mathbf{2}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{=}c\end{array}$

Hence the solution is $\mathbf{2}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{=}c$