Mumbai University > Electronics and Telecommunication > Sem 4 > Applied Maths 4

Marks: 5M

Year: May 2015

We have $f = \frac{\sqrt{1 + (y')^2}}{x}dx$

Since f does not contain any explicitly we shall use $\frac{\partial f}{\partial y'} = c$

$$\frac{\partial f}{\partial y'} = \frac{1}{x} \frac{\partial}{\partial y'} \bigg[1 + (y')^2\bigg]^{\frac{1}{2}} = \frac{1}{x} \frac{1}{2} \frac{2y'}{\sqrt{1 + (y')^2}}$$

But $\frac{\partial f}{\partial y'} = c$

$\therefore c = \frac{y'}{x\sqrt{1 + (y')^2}} \\ (cx)^2 (1 + (y')^2) = (y')^2 \\ \therefore c^2 x^2 + c^2 x^2 (y')^2 = (y')^2 \\ (y')^2 = \frac{cx}{1 - c^2 x^2} \\ y' = \frac{cx}{\sqrt{1 - c^2 x^2}}$

Now integrating both sides we get

$y = \int\frac{cx}{\sqrt{1 - c^2 x^2}}dx$

Let $1 - c^2 x^2 = t \therefore -2dx = dt$

$\therefore$ substituting we get

$y = -\frac{1}{2} \int \frac{cdt}{\sqrt{t}} \\ \therefore y = \frac{-c}{2} \frac{\sqrt{t}}{\frac{1}{2}} \\ \therefore y = -c\sqrt{t}$

Re-substituting the value we get,

$y = -c\sqrt{1 - c^2 x^2} + c_{2}$