Question

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

• A. 15
• B. 11
• C. 13
• D. 17

Answer 1

Answer: (A)

$y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
$\Rightarrow x\frac{dv}{dx}=\sqrt{v^2+16}$
$\Rightarrow \int \frac{dv}{\sqrt{v^2+16}}=\int \frac{dx}{x}$
$\Rightarrow \ln \left|v+\sqrt{v^2+16}\right|=\ln x+\ln C$
$\Rightarrow y+\sqrt{y^2+16x^2}=C{x}^2$
As $y(1)=3\Rightarrow C=8$
$\Rightarrow y(2)=15$