Let the solution curve of the differential eq | vedclass

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(A) Given the differential equation $x \frac{dy}{dx} - y = \sqrt{y^2 + 16x^2}$.Divide by $x$: $\frac{dy}{dx} - \frac{y}{x} = \sqrt{(\frac{y}{x})^2 + 1

Vedclass · Accepted Answer

$15$

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(A) Given the differential equation $x \frac{dy}{dx} - y = \sqrt{y^2 + 16x^2}$.Divide by $x$: $\frac{dy}{dx} - \frac{y}{x} = \sqrt{(\frac{y}{x})^2 + 16}$.Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.Substituting this into the equation: $v + x \frac{dv}{dx} - v = \sqrt{v^2 + 16}$.$x \frac{dv}{dx} = \sqrt{v^2 + 16} \Rightarrow \int \frac{dv}{\sqrt{v^2 + 16}} = \int \frac{dx}{x}$.Integrating both sides: $\ln|v + \sqrt{v^2 + 16}| = \ln|x| + \ln|C|$.$v + \sqrt{v^2 + 16} = Cx \Rightarrow \frac{y}{x} + \sqrt{\frac{y^2}{x^2} + 16} = Cx$.$y + \sqrt{y^2 + 16x^2} = Cx^2$.Using $y(1) = 3$: $3 + \sqrt{3^2 + 16(1)^2} = C(1)^2 \Rightarrow 3 + \sqrt{25} = C \Rightarrow C = 8$.Thus,$y + \sqrt{y^2 + 16x^2} = 8x^2$.For $x = 2$: $y + \sqrt{y^2 + 16(4)} = 8(4) = 32$.$y + \sqrt{y^2 + 64} = 32 \Rightarrow \sqrt{y^2 + 64} = 32 - y$.Squaring both sides: $y^2 + 64 = 1024 - 64y + y^2$.$64y = 960 \Rightarrow y = 15$.

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

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