Question

Let y = y(x), y>0, be a solution curve of the differential equation (1 + x²) dy=y (x–y)dx. If y(0)=1 and y(2√2) , = b then

• A. \(e β^{ − 1} = e^{-2} ( 3 + 2 √ 2 )\)
• B. \(e ^3 β^{ − 1} = e ( 3 + 2 √ 2 )\)
• C. \(e β^{ − 1} = e^{-2} ( 5+ √ 2 )\)
• D. \(e3 β^{ − 1} = e^{-2} ( 3+ 2√ 2 )\)

Answer 1

The Correct Option is D

To solve the differential equation given by:

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

We need to first rearrange the equation in a separable form:

\((1 + x^2) dy = y(x - y) dx\)

This can be rewritten as:

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

We can separate the variables as follows:

\(\frac{dy}{y} + \frac{y}{1 + x^2} \, dx = \frac{x}{1 + x^2} \, dx\)

Integrating both sides, we have:

\(\int \frac{dy}{y} + \int \frac{y}{1 + x^2} \, dx = \int \frac{x}{1 + x^2} \, dx\)

The solutions to these integrals are:

• \(\int \frac{dy}{y} = \ln |y|\)
• \int \frac{y}{1 + x^2} \, dx = \frac{y}{2} \ln(1 + x^2) (assuming y is constant with respect to x)
• \int \frac{x}{1 + x^2} \, dx = \frac{1}{2} \ln |1 + x^2|\)

Substituting into our equation, we get:

\(\ln |y| + \frac{y}{2} \ln(1 + x^2) = \frac{1}{2} \ln |1 + x^2| + C\)

Using the initial condition y(0) = 1, we find:

\(\ln |1| + \frac{1}{2} \ln(1) = \frac{1}{2} \ln |1| + C \Rightarrow C = 0\)

Simplified, the equation becomes:

\(\ln |y| = \frac{1}{2} \ln(1 + x^2)\)

Exponential both sides gives:

y = (1 + x^2)^{1/2}

Now, substitute x = 2\sqrt{2} and solve for b:

b = (1 + (2\sqrt{2})^2)^{1/2} = (1 + 8)^{1/2} = 3

Given the condition:

e^3 b^{-1} = e^{-2} (3 + 2\sqrt{2})

Compute the left-hand side:

e^3 \cdot (3)^{-1} = \frac{e^3}{3}

Thus, the correct option is verified:

e^3 \cdot b^{-1} = e^{-2} (3 + 2\sqrt{2})