Question

If the solution curve of the differential equation \(\frac{x + y - 2}{x + y - 1}\,\frac{dy}{dx} = \frac{x + y + 2}{x + y + 1}\), \(x + y > 2\), passes through the points \((\sqrt{2}, \sqrt{2})\) and \((2, \alpha)\), then \(2\alpha - \log_e\left(\frac{\alpha^2 + 4\alpha + 2}{6}\right)\) is equal to ________.

Hint:

Put \(v=x+y\) and separate variables; the integral of \(\dfrac{v}{v^2-2}\) gives a logarithm.

Answer 1

Correct Answer: 4

Let \(v=x+y\). Then \(\dfrac{dy}{dx}=\dfrac{(v+2)(v-1)}{(v+1)(v-2)}=\dfrac{v^2+v-2}{v^2-v-2}\), and \(\dfrac{dv}{dx}=1+\dfrac{dy}{dx}=\dfrac{2v^2-4}{v^2-v-2}\).
Separate: \(dx=\dfrac{v^2-v-2}{2(v^2-2)}\,dv=\dfrac12\left(1-\dfrac{v}{v^2-2}\right)dv\).
Integrate: \(x=\dfrac12\left[v-\dfrac12\ln(v^2-2)\right]+C\), i.e. \(2x=x+y-\dfrac12\ln\left((x+y)^2-2\right)+2C\), so \(y-x-\dfrac12\ln\left((x+y)^2-2\right)=K\).
Point \((\sqrt2,\sqrt2)\): \(v=2\sqrt2\), \(v^2-2=6\), so \(K=-\dfrac12\ln6\).
Point \((2,\alpha)\): \(\alpha-2-\dfrac12\ln\left((2+\alpha)^2-2\right)=-\dfrac12\ln6\Rightarrow\alpha-2=\dfrac12\ln\dfrac{\alpha^2+4\alpha+2}{6}\).
Hence \(2\alpha-4=\ln\dfrac{\alpha^2+4\alpha+2}{6}\), so \(2\alpha-\ln\dfrac{\alpha^2+4\alpha+2}{6}=4\).
Answer: 4