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

Marks: 6M

Year: May 2015

We have $f = (y')^2 + 12xy$

Since f contains x,y, & y' we must use $\frac{\partial f}{\partial y} - \frac{\partial}{\partial x} \bigg(\frac{\partial f}{\partial y'}\bigg) = 0 $ -> Enter equation

$\therefore \frac{\partial f}{\partial y} = 12x, \frac{\partial f}{\partial y'} = 2y'$

Hence the enter equation becomes

$12x - \frac{\partial}{\partial x}(2y') = 0 \\ \therefore 12x - \frac{2\partial^2 y}{\partial x^2} = 0 \\ \therefore \frac{\partial^2 y}{\partial x^2} = 6x$

Integrating we get $\frac{\partial y}{\partial x^2} = 3x^2 + c$

Integrating once again we get $y = x^3 + cx + c_{1}x$

Again by data y = 1, x = 1

1 + 1 = $c_{1} \therefore c_{1} = 0$

Hence the required curve is $y = x^3$