Let the solution curve of the differential equation x d y d x - y = y 2 + 16 x 2 , y 1 = 3 be y = y x . Then…

Given,

$x\frac{dy}{dx}-y=\sqrt{y^2+16x^2}$ 

$\Rightarrow \frac{dy}{dx}=\frac{\sqrt{y^2+16x^2}+y}{x}$

Let $y=xt\Rightarrow \frac{dy}{dx}=x\frac{dt}{dx}+t$

$\Rightarrow x\frac{dt}{dx}+t=t+\sqrt{t^2+16}$

$\Rightarrow \frac{dt}{\sqrt{t^2+16}}=\frac{dx}{x}$

Now integrating both side we get,

$\Rightarrow \int \frac{dt}{\sqrt{t^2+16}}=\int \frac{dx}{x}$

$\Rightarrow \ln \left|t+\sqrt{t^2+16}\right|=\ln x+\ln c$

$\Rightarrow \frac{y}{x}+\frac{\sqrt{y^2+16x^2}}{x}=cx$

$\because y\left(1\right)=3$ so $\Rightarrow \frac{3}{1}+\frac{\sqrt{3^2+16\times 1^2}}{1}=c\times 1\Rightarrow c=8$

So solution becomes,

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

Now $y\left(2\right)=\frac{y}{2}+\frac{\sqrt{y^2+16\times 2^2}}{2}=8\times 2$

$\Rightarrow y+\sqrt{y^2+64}=32$

$\Rightarrow y=15$