If a and b are arbitrary constants,then the d | vedclass

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(B) Given the family of curves $y = 	an(ax + b)$.First,differentiate with respect to $x$:$y_1 = \frac{dy}{dx} = \sec^2(ax + b) \cdot a$Since $\sec^2

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$(1 + y^2) y_2 - 2y y_1^2 = 0$

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(B) Given the family of curves $y = 	an(ax + b)$.First,differentiate with respect to $x$:$y_1 = \frac{dy}{dx} = \sec^2(ax + b) \cdot a$Since $\sec^2 	heta = 1 + 	an^2 	heta$,we have $y_1 = a(1 + y^2)$.Thus,$a = \frac{y_1}{1 + y^2}$.Now,differentiate $y_1 = a(1 + y^2)$ with respect to $x$ again:$y_2 = a(2y y_1) \cdot y_1 + a(1 + y^2) \cdot 0$ (Wait,differentiating $y_1 = a(1 + y^2)$ gives $y_2 = a(2y y_1)$).Substitute $a = \frac{y_1}{1 + y^2}$ into the equation $y_2 = 2ay y_1$:$y_2 = 2 \left( \frac{y_1}{1 + y^2} ight) y y_1$$y_2 = \frac{2y y_1^2}{1 + y^2}$Rearranging gives: $(1 + y^2) y_2 - 2y y_1^2 = 0$.

If $a$ and $b$ are arbitrary constants,then the differential equation corresponding to the family of curves $y = \tan(ax + b)$ is:

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