Question

$\large{\underline{\underline{\bf{Question}}}}$
Find the particular solution of the differential equation $\dfrac{dy}{dx} = -4x y^2$ given that y = 1, when x = 0.
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Answer 1

Solution -

If y ≠ 0, the given differential equation can be written as

• $\bf{\dfrac{dy}{y^2} = -4x\: dx}$

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Integrating both the sides

$\displaystyle\tt:\implies\: \: \: \: \: \: \: \: {\int \dfrac{dy}{y^2} = -4 \int x\: dx}$

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$\tt:\implies\: \: \: \: \: \: \: \: {- \dfrac{1}{y} = -2x^2 +C}$

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$\tt:\implies\: \: \: \: \: \: \: \: {\dfrac{1}{y} = 2x^2 - C}$

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$\tt:\implies\: \: \: \: \: \: \: \: {y = \dfrac{1}{2x^2 - C}}$⠀⠀...[1]

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Substituting y = 1 and x = 0 in [1]

$\tt:\implies\: \: \: \: \: \: \: \: {1 = \dfrac{1}{2(0)^2 - C}}$

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$\tt:\implies\: \: \: \: \: \: \: \: {1 = \dfrac{1}{2(0) - C}}$

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$\tt:\implies\: \: \: \: \: \: \: \: {1 = \dfrac{1}{-C}}$

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$\bf:\implies\: \: \: \: \: \: \: \: {C = -1}$

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Now, we will put the value of C in equation [1], to get the particular solution of the given differential equation.

$\tt:\implies\: \: \: \: \: \: \: \: {y = \dfrac{1}{2x^2 - (-1)}}$

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$\tt:\implies\: \: \: \: \: \: \: \: {y = \dfrac{1}{2x^2 + 1}}$

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★ $\underline{\sf{Hence,\: the\: required\: solution\: is}}$

• $\bf\purple{y = \dfrac{1}{2x^2 + 1}}$

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