Which of the following function is a solution of differential equation $y \cdot \frac{d y}{d x} = x {(\frac{d y}{d x})}^{2} + 1 ?$

y = A cos x + B sin x

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y = c x + \frac{1}{c}

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y = 2 x^{2} + 4

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y \cdot e x = c

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Solution

The correct option is D $y = c x + \frac{1}{c}$
Let $y = k x$ , where $k = c o n s t$
$∴ \frac{d y}{d x} = k$
Putting $\frac{d y}{d x}$ in the given equation:
$∴ y . (k) = x (k)^{2} + 1$
$\Rightarrow y = x k + \frac{1}{k}$
$∴ y = c x + \frac{1}{c}$ [Replacing k with c]

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Similar questions

y = c x + \frac{1}{x}

y \frac{d y}{d x} = x {(\frac{d y}{d x})}^{2} + 1

\frac{d^{2} y}{d x^{2}} + y = 0

y = A x + \frac{B}{x}

x^{2} \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} - y = 0

{(\frac{d y}{d x})}^{2} - x (\frac{d y}{d x}) + y = 0

y = a cos x + b sin x

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