Question

If the differential equation corresponding to the family of curves $y=c(x-c{)}^2,$ where $c$ is an arbitrary constant, is $8y^2=kxy\frac{dy}{dx}-{\left(\frac{dy}{dx}\right)}^3,$ then the value of $\frac{1}{3}\int \underset{t\to 0}{lim}\ln \left(\frac{1}{t}\underset{0}{\overset{t}{\int }}{(1+\frac{k}{2}\sin 3x)}^{k/x}dx\right)$ is

Answer 1

$y=c(x-c{)}^2\dots (1)\Rightarrow \frac{y}{x-c}=c(x-c)$
Differentiating equation $\left(1\right)$ w.r.t. $x,$
$y^{\prime }=2c(x-c)$
$\Rightarrow \frac{y^{\prime }}{2}=\frac{y}{x-c}\Rightarrow (x-c)=\frac{2y}{y^{\prime }}\Rightarrow c=x-\frac{2y}{y^{\prime }}$
Putting $c$ and $(x-c)$ values in equation $\left(1\right),$
$8y^2=4xy\frac{dy}{dx}-{\left(\frac{dy}{dx}\right)}^3$
$\Rightarrow k=4$


Let $L=\frac{1}{3}\underset{t\to 0}{lim}\ln \left(\frac{1}{t}\underset{0}{\overset{t}{\int }}{(1+\frac{k}{2}\sin 3x)}^{k/x}dx\right)$
$=\frac{1}{3}\ln \underset{t\to 0}{lim}\frac{\underset{0}{\overset{t}{\int }}(1+2\sin 3x{)}^{4/x}dx}{t}\left(\frac{0}{0}\text{form}\right)$
$=\frac{1}{3}\ln \underset{t\to 0}{lim}(1+2\sin 3t{)}^{4/t}$
(using L' Hospital's rule and Leibnitz’s theorem)
$=\ln e^{\frac{1}{3}\underset{t\to 0}{lim}\frac{4}{t}\left(2\sin 3t\right)}=\frac{1}{3}\underset{t\to 0}{lim}\frac{2\cdot 3\cdot 4\cdot \sin 3t}{3t}=8$