how to solve this?

[blamocur]

it say it correct, but it want me to express the answer in terms of the constant

for example if the differential equation is \(\displaystyle \frac{dy}{dx}\ln(\frac{dy}{dx}) = 5\), it want \(\displaystyle 5\) appear in the answer

That is answered by @mario99 in post # 16 and in post # 20, i.e., use Lambert's W function.

 

[logistic_guy]

thank you guys. i wached the video

It will help if you can show exactly what answer you gave, and exactly what "it" said in response, to make sure you aren't misinterpreting something, and that we know the details.

The mention of Lambert W in #18 may be relevant. Are you at all familiar with it? Have you looked it up?

blamocur's answer

\(\displaystyle y(x) = c_1x + c_2\)

That is answered by @mario99 in post # 16 and in post # 20, i.e., use Lambert's W function.

see what happen when i try to solve

\(\displaystyle y'(x) = c_1\)

\(\displaystyle \frac{dy}{dx}\ln(\frac{dy}{dx}) = c_1\ln(c_1) = c\)

\(\displaystyle \ln(c_1^{c_1}) = c\)

\(\displaystyle c_1^{c_1} = e^c\)

but the video say \(\displaystyle x^x = 5\)

my form isn't \(\displaystyle c_1^{c_1} = c\)

 

[blamocur]

You seem to be ignoring helpful suggestions, in particular the ones mentioning Lambert's function.

 

[Dr.Peterson]

blamocur's answer

\(\displaystyle y(x) = c_1x + c_2\)

So they want you to express your [imath]c_1[/imath] in terms of [imath]c[/imath].

It might still help if you answered my full question by saying what "it" said, word for word, in case there's more to it.

\(\displaystyle c_1^{c_1} = e^c\)

but the video say \(\displaystyle x^x = 5\)

my form isn't \(\displaystyle c_1^{c_1} = c\)

But it does have the form \(\displaystyle c_1^{c_1} = K\), where [imath]K=e^c[/imath]. Isn't that all you need? The name of the constant doesn't matter.

On the other hand, I myself didn't try for the form [imath]x^x[/imath]; I used the definition of W directly, by making another substitution.

Ultimately, if what you try doesn't work, just try a different path. You've been told it helps; so keep trying.