Question

If $(1+x^2)dy = y(y-x)dx$ and $y(1)=1$. Then $y(2√2)$ is:

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

Answer 1

Answer: (D)

${\sqrt2}$

Answer 2

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(x2 + y2) + 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:
y2 - 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.
**Answer. **D