(A) Given the slope of the curve is $\frac{dy}{dx} = 2xy$.Separating the variables,we get $\frac{dy}{y} = 2x \, dx$.Integrating both sides,we have $\i

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(A) Given the slope of the curve is $\frac{dy}{dx} = 2xy$.Separating the variables,we get $\frac{dy}{y} = 2x \, dx$.Integrating both sides,we have $\int \frac{dy}{y} = \int 2x \, dx$.This gives $\log y = x^{2} + C$.Since the curve passes through the point $(0,1)$,we substitute $x = 0$ and $y = 1$ into the equation:$\log(1) = (0)^{2} + C \implies 0 = 0 + C \implies C = 0$.Therefore,the equation of the curve is $\log y = x^{2}$.

Class 12 Mathematics Differential Equations Variable separable type differential equations

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The equation of the curve whose slope at any point is equal to $2xy$ and which passes through the point $(0,1)$ is

MHT CET 2020 All MHT CET

A
$\log y = x^{2}$
B
$\log y = \frac{1}{x}$
C
$\frac{1}{y} = x$
D
$\log y = x$

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