Question

Let the solution curve of the differential equation $x\frac{dy}{dx}-y=\sqrt{y^2+16x^2},y\left(1\right)=3$ be $y=y\left(x\right)$. Then $y\left(2\right)$ is equal to:

• A. $11$
• B. $17$
• C. $15$
• D. $15$

Answer 1

The correct option is C $15$

$x\frac{dy}{dx}-y=\sqrt{y^2+16x^2}$
$y=4x\tan \theta$
$\frac{dy}{dx}=4\tan \theta +4x{\sec }^2\theta \frac{d\theta }{dx}$
$4x\tan \theta +4x^2{\sec }^2\theta \frac{d\theta }{dx}-4x\tan \theta =4x\sec \theta$
$\int \sec \theta d\theta =\int \frac{dx}{x}$
$\log |\sec \theta +\tan \theta |=\log |x|+C$
$y\left(1\right)=3\Rightarrow 3=4\tan \theta$
$=\tan \theta =\frac{3}{4}\Rightarrow \sec \theta =\frac{5}{4}$
$\ln \left|\frac{8}{4}\right|=\ln |1|+C$
$\Rightarrow C=\ln 2$
$\therefore |\sec \theta +\tan \theta |=2|x|$
To find $y\left(2\right)$ put $x=2$
$\Rightarrow \tan \theta =\frac{y}{8}$
$\begin{array}{c}{(\sec \theta +\tan \theta )}^2=16\\ \sec \theta +\tan \theta =\pm 4\\ \sec \theta -\tan \theta =\pm \frac{1}{4}\\ 2\tan \theta =\frac{15}{4}=2\times \frac{y}{8}\end{array}$
$\Rightarrow \begin{array}{c}y=15\end{array}$