1. Given differential equation

$x\frac{dy}{dx}-y=\sqrt{y^2+16x^2}, \qquad y(1)=3$

We need to find $y(2)$.

2. Use the substitution

Since the equation contains $y$, $x$, and $x\dfrac{dy}{dx}-y$, divide by $x^2$ structure suggests putting

$y=vx$

where $v=v(x)$.

Then

$\frac{dy}{dx}=v+x\frac{dv}{dx}$

So,

$x\frac{dy}{dx}-y=x\left(v+x\frac{dv}{dx}\right)-vx=x^2\frac{dv}{dx}$

Also,

$y^2+16x^2=v^2x^2+16x^2=x^2(v^2+16)$

Hence,

$\sqrt{y^2+16x^2}=|x|\sqrt{v^2+16}$

Since the initial condition is at $x=1>0$ and we only need $x=2>0$, we take $x>0$, so $|x|=x$.

Thus the differential equation becomes

$x^2\frac{dv}{dx}=x\sqrt{v^2+16}$

or

$x\frac{dv}{dx}=\sqrt{v^2+16}$

So,

$\frac{dv}{\sqrt{v^2+16}}=\frac{dx}{x}$

3. Integrate both sides

Using

$\int \frac{dv}{\sqrt{v^2+a^2}}=\ln\left|v+\sqrt{v^2+a^2}\right|+C$

with $a=4$, we get

$\ln\left|v+\sqrt{v^2+16}\right|=\ln|x|+C$

Since $x>0$,

$v+\sqrt{v^2+16}=Cx$

4. Use the initial condition

Given $y(1)=3$, and $y=vx$, at $x=1$:

$v(1)=\frac{y(1)}{1}=3$

Substitute into the integrated equation:

$3+\sqrt{3^2+16}=C\cdot 1$

$3+\sqrt{25}=C$

$3+5=8$

So,

$C=8$

Hence,

$v+\sqrt{v^2+16}=8x$

5. Find $v$ when $x=2$

At $x=2$,

$v+\sqrt{v^2+16}=16$

So,

$\sqrt{v^2+16}=16-v$

Squaring both sides,

$v^2+16=(16-v)^2=256-32v+v^2$

$16=256-32v$

$32v=240$

$v=\frac{15}{2}$

Now,

$y=vx=\frac{15}{2}\cdot 2=15$

Thus,

$y(2)=15$

6. Check options

• A: $15$ ✅
• B: $11$
• C: $13$
• D: $17$

So the correct option is A.