Solve the first order linear differential equation

[chwala]

Homework Statement

$\dfrac{dy}{dx} + (\ln x) y = 6x$

Relevant Equations

integrating factor

I know that this is not elementary, but like thinking out of the box, how do we solve such... or which concepts do we draw from to reach somewhere...
I was able to understand up to this part,

Integrating factor = $e^{\int \ln x} = x\ln x-x$

 

[BvU]

Just so we don't waste time: is it $$\dfrac{dy}{dx} + \ln (x y) = 6x$$ or is it $$\dfrac{dy}{dx} + (\ln x) y = 6x\ ?$$

$\$

 

[julian]

....

Integral factor = $e^{\int \ln x} = x\ln x-x$

It should be $\int \ln x dx = x \ln x -x +C$, so $e^{\int \ln x dx} = K x^x e^{-x}$.

 

[pasmith]

Just so we don't waste time: is it $$\dfrac{dy}{dx} + \ln (x y) = 6x$$ or is it $$\dfrac{dy}{dx} + (\ln x) y = 6x\ ?$$

$\$

I would assume the first, since the second would have been written as y \ln x. But the first leads to a non-linear, non-separable equation and the second leads to a linear equation.

 

[chwala]

Just so we don't waste time: is it $$\dfrac{dy}{dx} + \ln (x y) = 6x$$ or is it $$\dfrac{dy}{dx} + (\ln x) y = 6x\ ?$$

$\$

Amended.

 

[julian]

I am well informed of this step, I seek the other steps after this...

Just to clarify — the expression you gave for the integrating factor wasn’t correct, so I was pointing that out.

 

[renormalize]

Homework Statement: $\dfrac{dy}{dx} + (\ln x) y = 6x$
Relevant Equations: integrating factor

From https://en.wikipedia.org/wiki/Integrating_factor:

Isn't your equation in exactly this form? Have you read the Wikipedia article?

 

[chwala]

I seek the steps to final solution specifically to this problem, that's the motivation behind the post. Of course, I am conversant with how to solve basic first order differential equations.

 

[chwala]

Just to clarify — the expression you gave for the integrating factor wasn’t correct, so I was pointing that out.

Wait, I thought when using integrating factor 'c' is silent or rather not considered... each term $y', P(x)$ and $Q(x)$ will be multiplied by the integrating factor, with no consideration to your c.

 

[renormalize]

I seek the steps to final solution specifically to this problem, that's the motivation behind the post. Of course, I am conversant with how to solve basic first order differential equations.

The point of my Wikipedia reference is to show you that solving your ODE using an integrating factor is just a matter of "turning the crank". Please show your step-by-step attempt to solve your equation to help us identify where you're stuck.

 

[chwala]

I thought i already mentioned the last step where I reached, ie finding the I.f ...my problem from that point is solely on how to use the integrating factor to this specifically arrive at the required solution.

 

[renormalize]

I thought i already mentioned the steps where I reached, finding the I.f ...my problem is solely on how to move towards solution.

As pointed out by @julian, you found the wrong integrating factor in your original post. Start over and show each of your steps. You should strive to accomplish the following:

1. Derive the proper integrating factor.
2. Display the ODE modified to include the integrating factor.
3. Integrate the modified ODE to display the solution $y(x)$, including an arbitrary constant of integration.

 

[chwala]

Where was it wrong? The missing of the constant $C$? My understanding is that contant $C$is embedded on the integral itself as the integrating factor will subsequently multiply out each term of the differential equation.

Just to clarify — the expression you gave for the integrating factor wasn’t correct, so I was pointing that out.

 

[renormalize]

Where was it wrong?

Per point 1) in my post #12, you should explicitly write the steps for deriving the integrating factor so we can check it.

 

[chwala]

I think we’re just going round in circles without any meaningful progress. In my mind, I’m very clear on where I had reached and where I am stuck. Getting the integrating factor is actually the least of my problems.

I already know how to form the integrating factor — and yes, I agree it’s a bit complex in this case because it involves integrating "ln x". But my real concern is how to move forward from there and actually solve the differential equation, not debating small technical points about the integrating factor.

Now, on the specific issue of whether the constant of integration — the "+ C" — should be included when forming the integrating factor: this is being misunderstood. When we find the integrating factor, we take the exponential of the integral of the function multiplying y. That process does not require including the constant.

Including the constant in the exponent just adds a constant multiple to the whole expression. That constant has no impact on the final solution because it gets absorbed into the constant of integration at the end. That’s why, as a standard method, we intentionally leave it out when calculating the integrating factor. This is not a mistake — it’s how the method is taught and used, including in exams and textbooks.

So if we keep focusing on whether "+ C" is there or not in the integrating factor, we’re missing the real issue. What I need help with is the next part of the process — how to proceed with solving once the integrating factor is applied. That’s where I am genuinely engaging with the math. The rest is just unnecessary back-and-forth over something that doesn’t affect the outcome.

 

[fresh_42]

By the time you wrote this, you could as well have typed your calculations, and we wouldn't argue in circles anymore, simply because it is unclear what exactly your problem is if you say you don't have one.

 

[renormalize]

So if we keep focusing on whether "+ C" is there or not in the integrating factor, we’re missing the real issue. What I need help with is the next part of the process — how to proceed with solving once the integrating factor is applied. That’s where I am genuinely engaging with the math. The rest is just unnecessary back-and-forth over something that doesn’t affect the outcome.

OK, so explicitly post here your integrating factor (without the integration constant) and we can go from there. If you're not willing to display at least some work effort for a homework problem, you can't really expect help here on Physics Forums.

 

[chwala]

OK, so explicitly post here your integrating factor (without the integration constant) and we can go from there. If you're not willing to display at least some work effort for a homework problem, you can't really expect help here on Physics Forums.

$e^{x \ln x - x} = x^x e^{-x}$.

 

[fresh_42]

$e^{x \ln x - x}$.

This isn't even an equation, let alone a solution. It is really hard to help you if you are so reluctant to provide information. Even your first post was ambiguous.

 

[chwala]

This isn't even an equation, let alone a solution. It is really hard to help you if you are so reluctant to provide information. Even your first post was ambiguous.

...explicitly post here your integrating factor...

@fresh_42 I would like help from this step, otherwise you may want to close this thread.
In post 1, I left out the 'e' typo error.

Thanks.

 

[renormalize]

$e^{x \ln x - x} = x^x e^{-x}$.

OK, great! Now multiply your original ODE by this and then try to rewrite all the terms involving $y$ and $y^{\prime}$ as the derivative of a single term.

 

[chwala]

OK, great! Now multiply your original ODE by this and then try to rewrite all the terms involving $y$ and $y^{\prime}$ as the derivative of a single term.

OK let me be specific, how do you integrate

$$ \dfrac{1}{x^x e^{-x}}\int
6x. x^x. e^{-x} dx $$

 

[renormalize]

OK let me be specific, how do you integrate
$$ \dfrac{1}{x^x e^{-x}}\int
6x. x^x. e^{-x} dx $$

You don't! The expression $y\left(x\right)=e^{x}x^{-x}\left(6\int e^{-x}x^{x+1}dx+k\right)$ (where $k$ is an arbitrary constant) represents the formal solution to the original ODE and that's as far as you can go. The integral exists but to my knowledge it has no known closed-form in terms of $x$, although I'm guessing you can expand the integrand in a Taylor series and integrate it term-by-term in closed form.

 

[chwala]

You don't! The expression $y\left(x\right)=e^{x}x^{-x}\left(6\int e^{-x}x^{x+1}dx+k\right)$ (where $k$ is an arbitrary constant) represents the formal solution to the original ODE and that's as far as you can go. The integral exists but to my knowledge it has no known closed-form in terms of $x$, although I'm guessing you can expand the integrand in a Taylor series and integrate it term-by-term in closed form.

That is exactly, where my problem is. Somehow this should be integrable... hence the reason of the post.

 

[DaveE]

Somehow this should be integrable

Wolfram Alpha says it's not...

 

[renormalize]

That is exactly, where my problem is. Somehow this should be integrable... hence the reason of the post.

But how do you know it should be integrable in closed form? There are an infinity of integrands that have no known closed-form antiderivatives.

 

[chwala]

You don't! The expression $y\left(x\right)=e^{x}x^{-x}\left(6\int e^{-x}x^{x+1}dx+k\right)$ (where $k$ is an arbitrary constant) represents the formal solution to the original ODE and that's as far as you can go. The integral exists but to my knowledge it has no known closed-form in terms of $x$, although I'm guessing you can expand the integrand in a Taylor series and integrate it term-by-term in closed form.

The formal solution to me does not look complete, but i fully understand the view of mathematicians coming to that conclusion.