Question

Dy/dx=x(lny). y(0)=1

Answer 1

Is it \(\ln(y(x))\) or \(\ln(x)\)?

Answer 2

No it's just like I wrote it I know I would do this
Dy/lny=xdx but then what

Answer 3

Integrate both sides. Here you would receive:
\[
\int \frac{1}{\ln(y)}dy=\int x\,dx\implies\\
\text{Li}(y(x))=\frac{x^2}{2}+C \implies\\
y(x)=\text{Li}^{-1}\left(\frac{x^2}{2}+C\right)
\]

Answer 4

Wait, change up the \(\text{Li}(x)\) to \(\text{li}(x)\)

Answer 5

(Sorry, I mixed up my offset log int)

Answer 6

Huh? Dy/lny

Answer 7

Yep. Integrating that gives you a function called the logarithmic integral, here:
en.wikipedia.org/wiki/Logarithmic_integral_function

Answer 8

Nope! It's not as hard as that.

Consider the behavior of the function in the neighborhood of the point \((0,1)\) (the initial value). \(\ln(1)=0\), so \(y'=0\). This means that in this neighborhood the function is approximately stationary. But, if there is no change in \(y\), then \(y=1\) and the function is stationary globally. By the uniqueness theorem, we must conclude that \(y(x)=1\) is the only solution. No special functions needed!

Answer 9

Check the answer!

if \(y(x)=1\),
\(y'(0)=0\)
\(x\ln(y(x))=x\ln(1)=0\)

Answer 10

@kmalone99