The particular solution of the differential e | vedclass

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Question

(A) Given differential equation: $x \, dy + 2y \, dx = 0$Rearranging the terms: $x \, dy = -2y \, dx$Separating variables: $\frac{dy}{2y} = -\frac{dx}

Vedclass · Accepted Answer

$x^2 y = 4$

Vedclass · Answer

(A) Given differential equation: $x \, dy + 2y \, dx = 0$Rearranging the terms: $x \, dy = -2y \, dx$Separating variables: $\frac{dy}{2y} = -\frac{dx}{x}$Integrating both sides: $\int \frac{1}{2y} \, dy = -\int \frac{1}{x} \, dx$$\frac{1}{2} \ln|y| = -\ln|x| + C$Multiply by $2$: $\ln|y| = -2\ln|x| + 2C$$\ln|y| = \ln|x^{-2}| + C_1$ (where $C_1 = 2C$)$y = e^{C_1} \cdot x^{-2} \implies y = \frac{K}{x^2}$ (where $K = e^{C_1}$)Given $x = 2$ and $y = 1$: $1 = \frac{K}{2^2} \implies K = 4$Thus, the particular solution is $y = \frac{4}{x^2}$, which simplifies to $x^2 y = 4$.

The particular solution of the differential equation $x \, dy + 2y \, dx = 0$, given that $y = 1$ when $x = 2$, is:

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