(C) Given that the slope $\frac{dy}{dx}$ is inversely proportional to $2y$,we have: $\frac{dy}{dx} = \frac{1}{2y}$ Separating the variables,we get: $2

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(C) Given that the slope $\frac{dy}{dx}$ is inversely proportional to $2y$,we have: $\frac{dy}{dx} = \frac{1}{2y}$ Separating the variables,we get: $2y \, dy = dx$ Integrating both sides: $\int 2y \, dy = \int dx$ $y^2 = x + C$ Since the curve passes through $(4, 3)$,we substitute $x = 4$ and $y = 3$: $(3)^2 = 4 + C$ $9 = 4 + C$ $C = 5$ Substituting the value of $C$ back into the equation,we get: $y^2 = x + 5$

Class 12 Mathematics Differential Equations Variable separable type differential equations

Medium

The slope of a curve at any point is inversely proportional to twice the $y$-coordinate of that point. If the curve passes through $(4, 3)$,then the equation of the curve is:

A
$x^2 = y + 5$
B
$y^2 = x - 5$
C
$y^2 = x + 5$
D
$x^2 = y - 5$

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Differential Equations

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Variable separable type differential equations

The general solution of ${x^2}\frac{{dy}}{{dx}} = 2$ is

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The particular solution of the differential equation $x dy + 2y dx = 0$,when $x = 2$ and $y = 1$ is

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

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Let $\alpha |x| = |y| e^{xy-\beta}$,where $\alpha, \beta \in \mathbb{N}$,be the solution of the differential equation $x dy - y dx + xy(x dy + y dx) = 0$ with the initial condition $y(1) = 2$. Then $\alpha + \beta$ is equal to:

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The slope of a curve at any point is inversely proportional to twice the $y$-coordinate of that point. If the curve passes through $(4, 3)$,then the equation of the curve is:

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The general solution of ${x^2}\frac{{dy}}{{dx}} = 2$ is

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Let $\alpha |x| = |y| e^{xy-\beta}$,where $\alpha, \beta \in \mathbb{N}$,be the solution of the differential equation $x dy - y dx + xy(x dy + y dx) = 0$ with the initial condition $y(1) = 2$. Then $\alpha + \beta$ is equal to:

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