Question

Let \(y = y(x), y > 0,\) be a solution curve of the differential equation \((1 +x^2)dy = y(x−y)dx.\) If \(y(0) = 1\) and \(y(2√2) = β\), then

• A. \(\frac{4}{\sqrt2}\)
• B. \(\frac{3}{\sqrt2}\)
• C. \(\frac{1}{\sqrt2}\)
• D. \({\sqrt2}\)

Answer 1

The Correct Option is D

We can solve the given differential equation using separation of variables.
\((1+x^2)dy = y(y-x)dx\)

Dividing both sides by y(y-x), we get:
\((\frac{1}{x} - \frac{1}{(x^2+y^2)}) dy - \frac{1}{y} dx = 0\)

Now, we can integrate both sides:
\(∫(\frac{1}{x} - \frac{1}{(x^2+y^2)}) dy - ∫\frac{1}{y} dx = C\)
where C is the constant of integration.

For the first integral, we can substitute \(u = x^2 + y^2, \frac{du}{dy} = 2y:\)
∫(\(\frac{1}{x}\) - \(\frac{1}{(x^2+Y^2)}\) dy = ∫\((\frac{1}{u})\) \((\frac{du}{dy})\) dy
= ∫(\(\frac{2y}{u}\)) dy
\(= ln|u| + K\)
\(= ln(x^2 + y^2) + K\)

For the second integral, we can directly integrate:
∫\(\frac{1}{y}\) dx = ln|y| + K'

Therefore, the equation becomes:
ln(x² + y²) - ln|y| = C'

Taking exponential of both sides, we get:
x² + y² = e^(C') |y|

Since \(y(1) = 1\), we have:
1² + 1² = e^(C') |1|
e^(C') = 2
C' = ln(2)

So, the equation becomes:
x² + y² = 2|y|

Substituting x = 2√2, we get:
(2√2)² + y² = 2|y|
8 + y² = 2|y|

Since y is positive, we can simplify this as:
\(y^2 - 2y + 8 = 0\)
Solving for y, we get:
\(y = 1 ± 3i\)

Since we want to find the value of y(2√2), we can substitute x = 2√2 in the equation x² + y² = 2|y| and solve for y:
(2√2)² + \(y^2\) = 2|y|
8 + \(y^2\) = 2|y|

Squaring both sides, we get:
64 + 16\(y^2\) +  y^4 = \(4 y^2\)
\(y^4\) + 12\(y^2\) - 64 = 0

Solving for \(y^2\), we get:
\(y^2\) = 4 or \(y^2\) = -16 (not possible since y is real)

So, we have:
y = ±2
Since y(1) = 1, we have y = 2.
Therefore, \(y(2√2) = 2\), so the answer is option (d) √2.