Question

Let $y = y(x)$ be the solution of the differential equation $\frac{dy}{dx} = 2x(x + y)^3 - x(x + y) - 1, \quad y(0) = 1.$Then, $\left( \frac{1}{\sqrt{2}} + y\left(\frac{1}{\sqrt{2}}\right) \right)^2$equals:

• A. $\frac{4}{4 + \sqrt{e}}$
• B. $\frac{3}{3 - \sqrt{e}}$
• C. $\frac{2}{1 + \sqrt{e}}$
• D. $\frac{1}{2 - \sqrt{e}}$

Answer 1

Answer: (D)

$\frac{1}{2 - \sqrt{e}}$

Answer 2

We are given the differential equation:

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

Let $x + y = t$. Therefore, we have:

$\frac{dt}{dx} = 2xt^3 - xt - 1$

This simplifies to:

$\frac{dt}{dx} = t^2 \quad \text{and} \quad \frac{dt}{dx} = x^2 \text{ for } x = 0$

Now solve the equation:

$\int \frac{dz}{2(2z - z)} = \int x dx$

After solving:

$\ln \left( \frac{z - 1}{z} \right) = x^2 + k$

Thus, $z = \frac{1}{2 - \sqrt{e}}$.