Question

Let \(y = y(x)\) be the solution of the differential equation \((x + 1)y' – y = e^{3x}(x + 1)^2\), with \(y(0)=\frac 13\). Then, the point \(x=−\frac 43\) for the curve \(y = y(x)\) is :

• A. not a critical point
• B. a point of local minima
• C. a point of local maxima
• D. a point of inflection

Answer 1

The Correct Option is B

We are given the differential equation:

((x + 1)y' – y = e^{3x}(x + 1)^2)

This can be rewritten as a linear first-order differential equation in the standard form:

y' + P(x)y = Q(x),

where P(x) = -\frac{1}{x+1} and Q(x) = e^{3x}(x + 1).

First, we need to find the integrating factor (IF), which is given by:

\text{IF} = e^{\int P(x) \, dx} = e^{\int -\frac{1}{x+1} \, dx}.

Calculating this integral, we have:

\int -\frac{1}{x+1} \, dx = -\ln|x+1|.

Therefore, the integrating factor is:

\text{IF} = e^{-\ln|x+1|} = \frac{1}{x+1}.

Multiplying the whole differential equation by the integrating factor:

\frac{1}{x+1}(y' + P(x)y) = \frac{1}{x+1}(Q(x))

becomes:

y' - \frac{1}{x+1}y = e^{3x}.

Now, the left side is the derivative of \frac{y}{x+1}, therefore:

\frac{d}{dx}\left(\frac{y}{x+1}\right) = e^{3x}.

Integrate both sides with respect to x:

\frac{y}{x+1} = \int e^{3x} \, dx = \frac{1}{3}e^{3x} + C,

where C is the constant of integration.

Thus, the solution is:

y = (x+1)\left(\frac{1}{3}e^{3x} + C\right).

We are given y(0) = \frac{1}{3} to find C. At x = 0:

\frac{1}{3} = (0+1)\left(\frac{1}{3} + C\right)

This gives C = 0.

Therefore, the particular solution is:

y = \frac{1}{3}e^{3x}(x+1).

To find the nature at x = -\frac{4}{3}, differentiate y:

y' = \frac{d}{dx}\left(\frac{1}{3}e^{3x}(x+1)\right)

= \frac{1}{3}\left(3e^{3x}(x+1) + e^{3x}\right).

Using y' = 0 to find critical points:

3(x+1)+1 = 0 \Rightarrow x = -\frac{4}{3}.

Differentiating again for y'' to classify the point, we find:

y'' = \ldots

Checking the sign of y'' at x = -\frac{4}{3}, we find it positive, which indicates a point of local minima.

Therefore, x = -\frac{4}{3} is indeed a point of local minima.