To reduce the differential equation frac{dy}{ | vedclass

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(B) The given equation is of the form $\frac{dy}{dx} + P(x)y = Q(x)y^n$,which is known as Bernoulli's equation.To linearize this equation,we divide bo

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$v = \frac{1}{y^{n-1}}$

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(B) The given equation is of the form $\frac{dy}{dx} + P(x)y = Q(x)y^n$,which is known as Bernoulli's equation.To linearize this equation,we divide both sides by $y^n$:$y^{-n} \frac{dy}{dx} + P(x)y^{1-n} = Q(x)$Now,we substitute $v = y^{1-n} = \frac{1}{y^{n-1}}$.Differentiating both sides with respect to $x$,we get:$\frac{dv}{dx} = (1-n)y^{-n} \frac{dy}{dx}$$\frac{1}{1-n} \frac{dv}{dx} = y^{-n} \frac{dy}{dx}$Substituting these into the equation,we get:$\frac{1}{1-n} \frac{dv}{dx} + P(x)v = Q(x)$Multiplying by $(1-n)$,we obtain:$\frac{dv}{dx} + (1-n)P(x)v = (1-n)Q(x)$This is a linear differential equation in $v$. Thus,the required substitution is $v = \frac{1}{y^{n-1}}$.

To reduce the differential equation $\frac{dy}{dx} + P(x)y = Q(x)y^n$ to the linear form,the substitution is

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