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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?

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Answered 2 years ago
Answered 2 years ago
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Since the initial condition is given at x=2x = 2, we obtain

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

So, the cofficient of yy' is 00 at x=2x=2. That indicates that something will likely go wrong.

Let's solve the given ODE.

(x2)dydx=y    dyy=dxx2,y0,  x2(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 yy, and the right side in relation to xx.

dyy=dxx2lny=lnx2+c\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

y=C(x2){\color{#4257b2}{y = C(x-2)}}

where C:=ecC:= e^c.

We can see that

y(2)=C(22)=01CRy(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)=yx2y' = f(x,y) = \frac{y}{x-2}

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

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