Question

If the solution $y = y(x)$ of the differential equation $\left( x^4 + 2x^3 + 3x^2 + 2x + 2 \right) dy - \left( 2x^2 + 2x + 3 \right) dx = 0$
satisfies $y(-1) = -\frac{\pi}{4}$, then $y(0)$ is equal to:

• A. $-\frac{\pi}{12}$
• B. 0
• C. $\frac{\pi}{4}$
• D. $\frac{\pi}{2}$

Answer 1

Answer: (C)

$\frac{\pi}{4}$

Answer 2

To solve this differential equation, separate the variables if possible and integrate both sides.

Rewrite the Differential Equation:
$(x^4 + 2x^3 + 3x^2 + 2x + 2) \, dy = (2x^2 + 2x + 3) \, dx$
Separation of Variables: Rewrite as:
$\frac{dy}{dx} = \frac{2x^2 + 2x + 3}{x^4 + 2x^3 + 3x^2 + 2x + 2}$

This equation may be complex to separate directly; therefore, assume an initial condition and use a direct integration or known solution pattern based on conditions $y(-1) = -\frac{\pi}{4}$ and evaluate at $x = 0$.

Using the Initial Condition $y(-1) = -\frac{\pi}{4}$:

By substituting values and integrating appropriately, we find:
$y(0) = \frac{\pi}{4}$.