Let y=y(x), x1, solve (x-1)dydx+2xy=1x-1, y(2)=1+e^42e^4.

Question

Let $y=y(x)$, $x>1$, solve $(x-1)\dfrac{dy}{dx}+2xy=\dfrac{1}{x-1}$, $y(2)=\dfrac{1+e^4}{2e^4}$. If $y(3)=\dfrac{e^\alpha+1}{\beta e^\alpha}$, then $\alpha+\beta$ is \_\_\_\_\_\_.

Answer: 14

Accepted Answer

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

Solving (I.F. involves $e^{2x}$-type factor since $\int \frac{2x}{x-1}dx = 2x+2\ln|x-1|$) and applying $y(2)$ gives $y(3)=\dfrac{e^6+1}{8e^6}$, i.e. $\alpha=6,\beta=8$, so $\alpha+\beta=14$.

Solutions · 1