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

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given the differential equation $\frac{dy}{dx} = -4xy^2$. Separating the variables,we get $\frac{dy}{y^2} = -4x dx$. Integrating both sides,we have $\int y^{-2} dy = -4 \int x dx$. This gives $-\frac{1}{y} = -4 \cdot \frac{x^2}{2} + C$,which simplifies to $-\frac{1}{y} = -2x^2 + C$. Multiplying by $-1$,we get $\frac{1}{y} = 2x^2 - C$. Given that $y = 1$ when $x = 0$,we substitute these values: $\frac{1}{1} = 2(0)^2 - C$,which implies $1 = -C$,so $C = -1$. Substituting $C = -1$ back into the equation $\frac{1}{y} = 2x^2 - C$,we get $\frac{1}{y} = 2x^2 - (-1) = 2x^2 + 1$. Therefore,the particular solution is $y = \frac{1}{2x^2 + 1}$.

Find the particular solution of the differential equation $\frac{dy}{dx} = -4xy^2$ given that $y = 1$ when $x = 0$.

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