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Question

# Does the initial value problem (x-2)y'=y, y(2)=1 have a solution? Does your result contradict our present theorems?

Solution

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Since the initial condition is given at $x = 2$, we obtain

$(2-2)y' = y \iff 0 \cdot y' = y.$

So, the cofficient of $y'$ is $0$ at $x=2$. That indicates that something will likely go wrong.

Let's solve the given ODE.

$(x-2)\frac{dy}{dx} = y \iff \frac{dy}{y} = \frac{dx}{x-2}, \quad y \neq 0,\; x \neq 2$

Now, the variables are separated. Integrate the left side in relation to $y$, and the right side in relation to $x$.

\begin{align*} \int \frac{dy}{y} &= \int \frac{dx}{x-2} \\ \ln |y| &= \ln |x-2| + c \end{align*}

By taking exponents, we obtain the general solution

${\color{#4257b2}{y = C(x-2)}}$

where $C:= e^c$.

We can see that

$y(2) = C(2-2) = 0 \neq 1 \quad \forall C \in \mathbb{R}$

However, this doesn't contradict our present theorems, because before applying them you need to write the equation in the standard form

$y' = f(x,y) = \frac{y}{x-2}$

Now, we can see that $f$ is not defined for $x=2$.

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