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

The Correct Option is D

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, \( z = \frac{1}{2 - \sqrt{e}} \).

Answer 2

Given the differential equation \( \frac{dy}{dx} = 2x(x + y)^3 - x(x + y) - 1 \) with \( y(0) = 1 \), we need to find \( \left( \frac{1}{\sqrt{2}} + y\left(\frac{1}{\sqrt{2}}\right) \right)^2 \).

Concept Used:

When a differential equation involves \( x+y \) repeatedly, a substitution \( t = x + y \) can simplify it. Then \( \frac{dt}{dx} = 1 + \frac{dy}{dx} \), which can be used to transform the equation into a separable or simpler form.

Step-by-Step Solution:

Step 1: Substitute \( t = x + y \). Then \( \frac{dt}{dx} = 1 + \frac{dy}{dx} \).

From the given equation:

\[ \frac{dy}{dx} = 2x t^3 - x t - 1 \]

So:

\[ \frac{dt}{dx} = 1 + \left( 2x t^3 - x t - 1 \right) = 2x t^3 - x t \] \[ \frac{dt}{dx} = x t (2 t^2 - 1) \]

Step 2: Separate variables and integrate.

\[ \frac{dt}{t(2 t^2 - 1)} = x \, dx \]

Factor the denominator: \( 2 t^2 - 1 = 2\left(t^2 - \frac12\right) \).
Use partial fractions for \( \frac{1}{t(2 t^2 - 1)} \).

Let \( \frac{1}{t(2 t^2 - 1)} = \frac{A}{t} + \frac{B t + C}{2 t^2 - 1} \).
Multiply through by \( t(2 t^2 - 1) \):

\[ 1 = A(2 t^2 - 1) + (B t + C) t = 2A t^2 - A + B t^2 + C t \] \[ 1 = (2A + B) t^2 + C t - A \]

Comparing coefficients:
\( 2A + B = 0 \) … (1)
\( C = 0 \) … (2)
\( -A = 1 \Rightarrow A = -1 \) … (3)
From (1): \( 2(-1) + B = 0 \Rightarrow B = 2 \).

Thus:

\[ \frac{1}{t(2 t^2 - 1)} = -\frac{1}{t} + \frac{2 t}{2 t^2 - 1} \]

Step 3: Integrate both sides.

\[ \int \left( -\frac{1}{t} + \frac{2 t}{2 t^2 - 1} \right) dt = \int x \, dx \] \[ - \ln |t| + \frac12 \ln |2 t^2 - 1| = \frac{x^2}{2} + C \]

Rewriting:

\[ \ln \left| \frac{\sqrt{2 t^2 - 1}}{t} \right| = \frac{x^2}{2} + C \] \[ \frac{\sqrt{2 t^2 - 1}}{t} = K e^{x^2 / 2}, \quad K = \pm e^C \]

Step 4: Use initial condition \( y(0) = 1 \) ⇒ \( t(0) = 0 + 1 = 1 \).

\[ \frac{\sqrt{2 (1)^2 - 1}}{1} = K e^{0} \Rightarrow \frac{\sqrt{1}}{1} = K \Rightarrow K = 1 \]

So:

\[ \frac{\sqrt{2 t^2 - 1}}{t} = e^{x^2 / 2} \] \[ \sqrt{2 t^2 - 1} = t e^{x^2 / 2} \]

Step 5: Square both sides:

\[ 2 t^2 - 1 = t^2 e^{x^2} \] \[ 2 t^2 - t^2 e^{x^2} = 1 \] \[ t^2 (2 - e^{x^2}) = 1 \] \[ t^2 = \frac{1}{2 - e^{x^2}} \]

Recall \( t = x + y \).

Step 6: Evaluate \( \left( \frac{1}{\sqrt{2}} + y\left(\frac{1}{\sqrt{2}}\right) \right)^2 \).

Let \( x_0 = \frac{1}{\sqrt{2}} \), then \( t_0 = x_0 + y(x_0) \).

\[ t_0^2 = \frac{1}{2 - e^{x_0^2}} = \frac{1}{2 - e^{1/2}} \]

So:

\[ \left( \frac{1}{\sqrt{2}} + y\left(\frac{1}{\sqrt{2}}\right) \right)^2 = t_0^2 = \frac{1}{2 - \sqrt{e}} \]

Therefore, the required value is \( \mathbf{\frac{1}{2 - \sqrt{e}}} \).