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Question

(B) Given the differential equation $\frac{dy}{dx} = -4xy^2$. Separating the variables,we get $\frac{dy}{y^2} = -4x \, dx$. Integrating both sides,$\i

Vedclass · Accepted Answer

$y = \frac{1}{2x^2 + 1}$

Vedclass · Answer

(B) Given the differential equation $\frac{dy}{dx} = -4xy^2$. Separating the variables,we get $\frac{dy}{y^2} = -4x \, dx$. Integrating both sides,$\int y^{-2} \, dy = \int -4x \, dx$. This gives $-\frac{1}{y} = -2x^2 + C$,or $\frac{1}{y} = 2x^2 - C$. Using the initial condition $x = 0, y = 1$: $\frac{1}{1} = 2(0)^2 - C \implies 1 = -C \implies C = -1$. Substituting $C = -1$ into the equation $\frac{1}{y} = 2x^2 - C$,we get $\frac{1}{y} = 2x^2 + 1$. Therefore,$y = \frac{1}{2x^2 + 1}$.

The particular solution of the differential equation $\frac{dy}{dx} = -4xy^2$ with the initial condition $x = 0, y = 1$ is . . . . . . .

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