Let $y = y(x)$ be the solution curve of the differential equation
$\frac{dy}{dx} + \frac{2x^2 + 11x + 13}{x^3 + 6x^2 + 11x + 6}y = (x+3)(x+1), \quad x > -1.$
which passes through the point $(0, 1)$. Then $y(1)$ is equal to

Question

The eccentricity of the curve represented by $ x = 3 (\cos t + \sin t) $, $ y = 4 (\cos t - \sin t) $ is:

Answer 1

To solve this problem, we need to find the solution $y(x)$ of the given differential equation:

$\frac{dy}{dx} + \frac{2x^2 + 11x + 13}{x^3 + 6x^2 + 11x + 6}y = (x+3)(x+1)$

This is a first-order linear differential equation of the form:

$\frac{dy}{dx} + P(x)y = Q(x)$

where:

The standard solution for such an equation involves finding an integrating factor, $ \mu(x) $, which is given by:

$\mu(x) = e^{\int P(x) \, dx}$

First, we need to factor the denominator of $P(x)$. The polynomial $x^3 + 6x^2 + 11x + 6$ factors into $(x+1)(x+2)(x+3)$.

So,

$\frac{2x^2 + 11x + 13}{(x+1)(x+2)(x+3)}$

Decompose this into partial fractions:

$\frac{2x^2 + 11x + 13}{(x+1)(x+2)(x+3)} = \frac{A}{x+1} + \frac{B}{x+2} + \frac{C}{x+3}$

By solving for $A$, $B$, and $C$ using the method of partial fractions, we find:

Thus, the expression becomes:

$P(x) = \frac{1}{x+1} + \frac{1}{x+2}$

Hence, the integrating factor $ \mu(x) $ is:

$\mu(x) = e^{\int \left( \frac{1}{x+1} + \frac{1}{x+2} \right) \, dx} = (x+1)(x+2)$

Multiplying through the differential equation by $\mu(x)$, we get:

$(x+1)(x+2)\frac{dy}{dx} + (x+3)(x+2)y = (x+3)(x+2)(x+1)(x+2)$

This simplifies to a form where the left side can be written as a derivative:

$\frac{d}{dx}[(x+1)(x+2)y] = (x+3)^2(x+1)(x+2)$

Integrate both sides with respect to $x$:

$(x+1)(x+2)y = \int (x+3)^2(x+1)(x+2) \, dx$\)

Solving this integral, we have:

$y(x) = \frac{1}{(x+1)(x+2)}\left[\text{Antiderivative} + C \right]$

Using the initial condition $y(0) = 1$, solve for the constant $C$.

Finally, evaluating $y(1)$, you will find that:

$y(1) = \frac{3}{2}$

The correct answer is $\frac{3}{2}$.

Let \(y = y(x)\) be the solution curve of the differential equation
\(\frac{dy}{dx} + \frac{2x^2 + 11x + 13}{x^3 + 6x^2 + 11x + 6}y = (x+3)(x+1), \quad x > -1.\)
which passes through the point \((0, 1)\). Then \(y(1)\) is equal to