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Differential equation

Need help with question 3.

Not entirely sure how to do it do i need to find an integrating factor?
Original post by Super199
Need help with question 3.

Not entirely sure how to do it do i need to find an integrating factor?


Might be helpful to take a picture of, LaTeX\LaTeX or link to the question. :biggrin:
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Super199
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Original post by Kvothe the arcane
Might be helpful to take a picture of, LaTeX\LaTeX or link to the question. :biggrin:

Well i tried uploading it but it wont let me so i shall type it out.

3. Find the general solution of the differential equation dy/dx-3x^2y= xe^(x^3), giving y explicitly in terms of x. Find also the particular solution for which y=1 when x=0
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Original post by Super199
Need help with question 3.

Not entirely sure how to do it do i need to find an integrating factor?


where is the question
Original post by Super199
Well i tried uploading it but it wont let me so i shall type it out.

3. Find the general solution of the differential equation dy/dx-3x^2y= xe^(x^3), giving y explicitly in terms of x. Find also the particular solution for which y=1 when x=0


To your question, yes.

The integrating factor is e∫−3x2dx\displaystyle e^{\int -3x^2 dx}

Work it out and the equation by it.
(edited 8 years ago)
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Super199
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Original post by Kvothe the arcane
To your question, yes.

The integrating factor is e∫−3x2dx\displaystyle e^{\int -3x^2 dx}

Work it out and the equation by it.

If i could upload a picture it would help but i got the integrating factor as e^-x^3.

Multiplied through :

Dy/dxe^-x^3 - 3x^2ye^-x^3= x

D/dx(ye^-x^3)=x

Ye^-x^3= x^2/2+c.
Im having troubles with e. If you divide by e^(-x^3) do you get y= (x^2+c)e^x^3
Original post by Super199
If i could upload a picture it would help but i got the integrating factor as e^-x^3.

Multiplied through :

Dy/dxe^-x^3 - 3x^2ye^-x^3= x

D/dx(ye^-x^3)=x

Ye^-x^3= x^2/2+c.
Im having troubles with e. If you divide by e^(-x^3) do you get y= (x^2+c)e^x^3

So you have ye−x3=x22+Cye^{-x^3}=\dfrac{x^2}{2}+C
Perhaps rewrite as yex3=x22+C\dfrac{y}{e^{x^3}}=\dfrac{x^2}{2}+C?
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Super199
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Original post by Kvothe the arcane
So you have ye−x3=x22+Cye^{-x^3}=\dfrac{x^2}{2}+C
Perhaps rewrite as yex3=x22+C\dfrac{y}{e^{x^3}}=\dfrac{x^2}{2}+C?


Got it cheers need help with another if you dont mind. Im having trouble with the integrating factor again.

Dy/dx +y/(xlnx)= 1/x. With y=1 when x=e is to be solved using an integrating factor. Find the integrating factor and solve the differential equation. Sketch the solution curve for x>1.
Integral of xlnx im not sure. Do i have to use parts? But the problem is it is (xlnx)^-1. So not sure. Or can i write it as 1/x*lnx. Idk cheers
Original post by Super199
Got it cheers need help with another if you dont mind. Im having trouble with the integrating factor again.

Dy/dx +y/(xlnx)= 1/x. With y=1 when x=e is to be solved using an integrating factor. Find the integrating factor and solve the differential equation. Sketch the solution curve for x>1.
Integral of xlnx im not sure. Do i have to use parts? But the problem is it is (xlnx)^-1. So not sure. Or can i write it as 1/x*lnx. Idk cheers

Had to think about it.

But it might be helpful to recall that ddxlnx=1x\dfrac{d}{dx}lnx=\dfrac{1}{x} and ∫f′(x)f(x)dx=ln(f(x))+C\displaystyle \int \dfrac{f'(x)}{f(x)}dx=ln(f(x))+C.

Not IBP, no.
(edited 8 years ago)
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Super199
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Original post by Kvothe the arcane
Had to think about it.

But it might be helpful to recall that ddxlnx=1x\dfrac{d}{dx}lnx=\dfrac{1}{x} and ∫f′(x)f(x)dx=ln(f(x))+C\displaystyle \int \dfrac{f'(x)}{f(x)}dx=ln(f(x))+C.

Not IBP, no.

I cant see the link sorry. I know that lnx integrates to xlnx-x and also what you said. But I cant see the link
Original post by Super199
I cant see the link sorry. I know that lnx integrates to xlnx-x and also what you said. But I cant see the link


1xlnx=1x×1lnx\dfrac{1}{xlnx}=\dfrac{1}{x} \times \dfrac{1}{lnx}

What is the relationship between 1/x and lnx? And how might that be related to what I wrote above?
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Zacken
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Original post by Super199
Got it cheers need help with another if you dont mind. Im having trouble with the integrating factor again.

Dy/dx +y/(xlnx)= 1/x. With y=1 when x=e is to be solved using an integrating factor. Find the integrating factor and solve the differential equation. Sketch the solution curve for x>1.
Integral of xlnx im not sure. Do i have to use parts? But the problem is it is (xlnx)^-1. So not sure. Or can i write it as 1/x*lnx. Idk cheers


Use the substitution u=ln⁡xu = \ln x, you'll end up with ∫duu\int\frac{\mathrm{d}u}{u}.
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Super199
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Original post by Zacken
Use the substitution u=ln⁡xu = \ln x, you'll end up with ∫duu\int\frac{\mathrm{d}u}{u}.


Right that helps now how do you integrate lnx/x
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Super199
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Original post by Super199
Right that helps now how do you integrate lnx/x


Not to worry got it. Using parts
Reply 14
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Zacken
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Original post by Super199
Right that helps now how do you integrate lnx/x


I'm not sure what you're asking. You want to integrate ∫dxxln⁡x\int \frac{\mathrm{d}x}{x \ln x} , right? That's ln⁡(ln⁡x)\ln (\ln x).

Integrating ln⁡xx\displaystyle \frac{\ln x}{x} is an easy one if you use u=ln⁡xu = \ln x you should get ∫u du\int u \, \mathrm{d}u

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