The particular solution of the differential e | vedclass

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Question

(A) Step $1$: Rearrange the differential equation: $x \, dy = -2y \, dx$.Step $2$: Separate the variables: $\frac{dy}{y} = -2 \frac{dx}{x}$.Step $3$:

Vedclass · Accepted Answer

$x^2 y = 4$

Vedclass · Answer

(A) Step $1$: Rearrange the differential equation: $x \, dy = -2y \, dx$.Step $2$: Separate the variables: $\frac{dy}{y} = -2 \frac{dx}{x}$.Step $3$: Integrate both sides: $\int \frac{dy}{y} = -2 \int \frac{dx}{x} \implies \ln|y| = -2 \ln|x| + C$.Step $4$: Simplify: $\ln|y| = \ln|x^{-2}| + C \implies \ln|y| = \ln|\frac{1}{x^2}| + C \implies y = \frac{k}{x^2}$, where $k = e^C$.Step $5$: Apply the initial condition $x = 2, y = 1$: $1 = \frac{k}{2^2} \implies 1 = \frac{k}{4} \implies k = 4$.Step $6$: 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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