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(B) The given differential equation is $(x+1) \frac{dy}{dx} - y = e^{3x}(x+1)^2$.Dividing by $(x+1)^2$,we get $\frac{(x+1) \frac{dy}{dx} - y}{(x+1)^2}

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a point of local minima

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(B) The given differential equation is $(x+1) \frac{dy}{dx} - y = e^{3x}(x+1)^2$.Dividing by $(x+1)^2$,we get $\frac{(x+1) \frac{dy}{dx} - y}{(x+1)^2} = e^{3x}$.This can be written as $\frac{d}{dx} \left( \frac{y}{x+1} \right) = e^{3x}$.Integrating both sides with respect to $x$,we get $\frac{y}{x+1} = \int e^{3x} dx = \frac{e^{3x}}{3} + C$.Given $y(0) = \frac{1}{3}$,substituting $x=0$ and $y=\frac{1}{3}$ gives $\frac{1/3}{1} = \frac{e^0}{3} + C$,so $\frac{1}{3} = \frac{1}{3} + C$,which implies $C=0$.Thus,$y = \frac{(x+1)e^{3x}}{3}$.To find the critical points,we find $\frac{dy}{dx} = \frac{1}{3} [ (x+1) \cdot 3e^{3x} + e^{3x} \cdot 1 ] = \frac{e^{3x}}{3} [ 3x + 3 + 1 ] = \frac{e^{3x}}{3} (3x+4)$.Setting $\frac{dy}{dx} = 0$,we get $3x+4=0$,so $x = -\frac{4}{3}$.For $x  -\frac{4}{3}$,$\frac{dy}{dx} > 0$.Since the derivative changes sign from negative to positive at $x = -\frac{4}{3}$,it is a point of local minima.

Let $y=y(x)$ be the solution of the differential equation $(x+1) y^{\prime}-y=e^{3 x}(x+1)^{2}$,with $y(0)=\frac{1}{3}$. Then,the point $x=-\frac{4}{3}$ for the curve $y = y ( x )$ is

Similar Questions

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$R$. Integrating factor of $\frac{dy}{dx} + P(x)y = Q(x)$ is $e^{\int P(x) dx}$.
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