## Related questions with answers

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

Solution

VerifiedSince 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 $x=2$.

## Create a free account to view solutions

## Create a free account to view solutions

## Recommended textbook solutions

#### Advanced Engineering Mathematics

10th Edition•ISBN: 9780470458365 (5 more)Erwin Kreyszig#### Advanced Engineering Mathematics

9th Edition•ISBN: 9780471488859 (3 more)Erwin Kreyszig#### Advanced Engineering Mathematics

6th Edition•ISBN: 9781284105902 (2 more)Dennis G. Zill## More related questions

1/4

1/7