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Step 1: Express the Differential Equation Clearly
The given differential equation is:
$ (1 + x^2)\,dy = y(x - y)\,dx. $
Rewriting in the derivative form:
$ \displaystyle \frac{dy}{dx} = \frac{y(x - y)}{1 + x^2}. $
The initial conditions are:
$ y(0) = 1 $ and $ y(2\sqrt{2}) = \beta. $
Step 2: Rewrite and Look for a Suitable Substitution
Notice that:
$ \displaystyle \frac{dy}{dx} = \frac{yx - y^2}{1 + x^2}. $
We suspect a substitution involving $ \frac{1}{y} $ might simplify the expression, since it contains terms like $ yx $ and $ y^2 $ in the numerator.
Step 3: Substitution $ t = \frac{1}{y} $
Let
$ \displaystyle t = \frac{1}{y}. $
Then,
$ \displaystyle \frac{dt}{dx} = -\frac{1}{y^2}\,\frac{dy}{dx} =
-y^{-2}\,\frac{dy}{dx}. $
Since $ y = \frac{1}{t}, $
$ \displaystyle y^2 = \frac{1}{t^2}. $
Therefore,
$ \displaystyle \frac{dy}{dx} = -\frac{1}{t^2}\,\frac{dt}{dx}. $
Step 4: Transform the Original Differential Equation in Terms of $ t $
Substitute $ y = \frac{1}{t} $ and
$ \frac{dy}{dx} = -\frac{1}{t^2}\,\frac{dt}{dx} $
into
$ \displaystyle \frac{dy}{dx} = \frac{y(x - y)}{1 + x^2}. $
We get:
$ \displaystyle -\frac{1}{t^2}\,\frac{dt}{dx} = \frac{\tfrac{1}{t}x - \tfrac{1}{t^2}}{1 + x^2}. $
Multiply through by $ -t^2 $:
$ \displaystyle \frac{dt}{dx} = \frac{tx - 1}{1 + x^2}. $
This rearranges to:
$ \displaystyle \frac{dt}{dx} + t\,\frac{x}{1 + x^2} = \frac{1}{1 + x^2}. $
Step 5: Identify the Integrating Factor
We have a first-order linear differential equation in $ t $:
$ \displaystyle \frac{dt}{dx} + p(x)\,t = q(x), $
where
$ p(x) = \frac{x}{1 + x^2} $
and
$ q(x) = \frac{1}{1 + x^2}. $
The integrating factor (I.F.) is
$ \displaystyle e^{\int p(x)\,dx} = e^{\int \frac{x}{1 + x^2}\,dx}. $
This integral is:
$ \displaystyle \int \frac{x}{1 + x^2}\,dx = \frac{1}{2}\ln\bigl(1 + x^2\bigr). $
Hence, the integrating factor is:
$ \displaystyle e^{\tfrac{1}{2} \ln(1 + x^2)} = (1 + x^2)^{\tfrac{1}{2}} = \sqrt{1 + x^2}. $
Step 6: Multiply Through by the Integrating Factor
Multiplying
$ \displaystyle \frac{dt}{dx} + t\,\frac{x}{1 + x^2} = \frac{1}{1 + x^2} $
by
$ \sqrt{1 + x^2}, $
we get:
$ \displaystyle \sqrt{1 + x^2}\,\frac{dt}{dx} + t\,\frac{x\,\sqrt{1 + x^2}}{1 + x^2} = \frac{\sqrt{1 + x^2}}{1 + x^2}. $
The left-hand side is exactly the derivative of
$ t\,\sqrt{1 + x^2}. $
Hence we can write:
$ \displaystyle \frac{d}{dx}\Bigl(t\,\sqrt{1 + x^2}\Bigr) = \frac{\sqrt{1 + x^2}}{1 + x^2}. $
Observe that
$ \displaystyle \frac{\sqrt{1 + x^2}}{1 + x^2} = \frac{1}{\sqrt{1 + x^2}}. $
Step 7: Integrate Both Sides
We integrate each side with respect to $ x $:
$ \displaystyle t\,\sqrt{1 + x^2} = \int \frac{1}{\sqrt{1 + x^2}}\,dx + C. $
We know
$ \displaystyle \int \frac{1}{\sqrt{1 + x^2}}\,dx = \ln\bigl(x + \sqrt{x^2 + 1}\bigr). $
Thus:
$ \displaystyle t\,\sqrt{1 + x^2} = \ln\bigl(x + \sqrt{x^2 + 1}\bigr) + C. $
Recalling $ t = \frac{1}{y}, $ we finally get:
$ \displaystyle \frac{\sqrt{1 + x^2}}{y} = \ln\bigl(x + \sqrt{x^2 + 1}\bigr) + C. $
Step 8: Apply the Condition $ y(0) = 1 $ to Find $ C $
When $ x=0, $ we have $ y(0)=1. $ Substitute into
$ \displaystyle \frac{\sqrt{1 + 0^2}}{1} = \ln(0 + \sqrt{0 + 1}) + C. $
Hence:
$ \displaystyle 1 = \ln(1) + C. $
Since $ \ln(1)=0, $ we get
$ C=1. $
The particular solution becomes:
$ \displaystyle \frac{\sqrt{1 + x^2}}{y} = \ln\bigl(x + \sqrt{x^2 + 1}\bigr) + 1. $
Step 9: Use the Condition $ y(2\sqrt{2}) = \beta $
Now, let $ x=2\sqrt{2} $ and $ y(2\sqrt{2}) = \beta. $ Then
$ \sqrt{1 + (2\sqrt{2})^2} = \sqrt{1 + 8} = 3. $
Substitute into the solution:
$ \displaystyle \frac{3}{\beta} = \ln\bigl(2\sqrt{2} + 3\bigr) + 1. $
Rearranging:
$ \displaystyle 3\,\beta^{-1} = 1 + \ln\bigl(3 + 2\sqrt{2}\bigr). $
Step 10: Exponentiate to Obtain the Final Relation
Exponentiating both sides:
$ \displaystyle e^{3\,\beta^{-1}} = e^{1 + \ln(3 + 2\sqrt{2})}. $
Using properties of exponents:
$ \displaystyle e^{a + b} = e^a \cdot e^b, $
we get:
$ \displaystyle e^{3\,\beta^{-1}} = e \times e^{\ln\bigl(3 + 2\sqrt{2}\bigr)} = e\,\bigl(3 + 2\sqrt{2}\bigr). $
This matches the correct answer:
$ \displaystyle e^{3 \beta^{-1}} = e\,(3 + 2\sqrt{2}). $
