The particular solution of frac{y}{x} frac{dy | vedclass

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(D) Given the differential equation: $\frac{y}{x} \frac{dy}{dx} = \frac{1+y^2}{1+x^2}$ Separating the variables,we get: $\frac{y}{1+y^2} dy = \frac{x}

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$5\left(1+y^2\right)=2\left(1+x^2\right)$

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(D) Given the differential equation: $\frac{y}{x} \frac{dy}{dx} = \frac{1+y^2}{1+x^2}$ Separating the variables,we get: $\frac{y}{1+y^2} dy = \frac{x}{1+x^2} dx$ Integrating both sides: $\int \frac{y}{1+y^2} dy = \int \frac{x}{1+x^2} dx$ Multiply by $2$ on both sides to facilitate integration: $\int \frac{2y}{1+y^2} dy = \int \frac{2x}{1+x^2} dx$ This results in: $\ln|1+y^2| = \ln|1+x^2| + \ln C$ Using logarithmic properties: $\ln(1+y^2) = \ln(C(1+x^2))$ Thus: $1+y^2 = C(1+x^2)$ Given $x=2$ and $y=1$: $1+(1)^2 = C(1+(2)^2) \Rightarrow 2 = 5C \Rightarrow C = \frac{2}{5}$ Substituting $C$ back into the equation: $1+y^2 = \frac{2}{5}(1+x^2)$ Multiplying by $5$: $5(1+y^2) = 2(1+x^2)$

The particular solution of $\frac{y}{x} \frac{dy}{dx} = \frac{1+y^2}{1+x^2}$ when $x=2, y=1$ is

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