Let the solution curve of the differential eq | vedclass

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Question

(A) Given the differential equation: $x dy - y dx = \sqrt{x^{2} + y^{2}} dx$. Divide both sides by $x^{2}$ (since $x > 0$): $\frac{x dy - y dx}{x^{2}}

Vedclass · Accepted Answer

$4$

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(A) Given the differential equation: $x dy - y dx = \sqrt{x^{2} + y^{2}} dx$. Divide both sides by $x^{2}$ (since $x > 0$): $\frac{x dy - y dx}{x^{2}} = \frac{\sqrt{x^{2} + y^{2}}}{x^{2}} dx$. This simplifies to: $d\left(\frac{y}{x}\right) = \sqrt{1 + \left(\frac{y}{x}\right)^{2}} \cdot \frac{1}{x} dx$. Integrating both sides: $\int \frac{d(\frac{y}{x})}{\sqrt{1 + (\frac{y}{x})^{2}}} = \int \frac{1}{x} dx$. Using the standard integral $\int \frac{du}{\sqrt{1 + u^{2}}} = \ln|u + \sqrt{1 + u^{2}}| + C$, we get: $\ln\left(\frac{y}{x} + \sqrt{1 + \frac{y^{2}}{x^{2}}}\right) = \ln x + C$. Given $y(1) = 0$, substitute $x = 1, y = 0$: $\ln(0 + \sqrt{1 + 0}) = \ln(1) + C \Rightarrow 0 = 0 + C \Rightarrow C = 0$. Thus, $\frac{y}{x} + \sqrt{1 + \frac{y^{2}}{x^{2}}} = x$. Multiplying by $x$: $y + \sqrt{x^{2} + y^{2}} = x^{2}$. To find $y(3)$, substitute $x = 3$: $y + \sqrt{9 + y^{2}} = 9$. $\sqrt{9 + y^{2}} = 9 - y$. Squaring both sides: $9 + y^{2} = 81 - 18y + y^{2}$. $18y = 72 \Rightarrow y = 4$.

Let the solution curve of the differential equation $x dy - y dx = \sqrt{x^{2} + y^{2}} dx$, where $x > 0$ and $y(1) = 0$, be $y = y(x)$. Then $y(3)$ is equal to:

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