- #1

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[tex]y' + y/(x+1) = x^2[/tex]

My first instinct with 1st order DEs is to seperate the variables.

But I can't.

Help.

Edit: that's "y/(x+1)" not "(y/x)+1". My LaTeX is rusty

- Thread starter Gwilim
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- #1

- 126

- 0

[tex]y' + y/(x+1) = x^2[/tex]

My first instinct with 1st order DEs is to seperate the variables.

But I can't.

Help.

Edit: that's "y/(x+1)" not "(y/x)+1". My LaTeX is rusty

- #2

Cyosis

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Sorry for asking a question? Anyway what do you know about integrating factors? Multiply the differential equation by u(x) first then try to write the LHS as d(uy)/dx.

Last edited:

- #3

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So the integrating factor is e^ln(x+1)=x+1

Multiplying both sides gives me [tex](x+1)y' + y = (x+1)x^2[/tex]

And that's where I get stuck.

Is there a reasonably short guide to this anywhere?

- #4

Cyosis

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You are pretty close. I am sure you can find a guide somewhere, but I have not much to do currently so I'll try to help you out.

Lets multiply the differential equation by u(x) this yields [itex]u(x)y'(x)+\frac{y(x)u(x)}{x+1}=u(x)x^2[/itex]. We want to write the left hand side as [itex]d(uy)/dx[/itex].

[tex]

\frac{duy}{dx}=u \frac{dy}{dx}+y\frac{du}{dx}

[/tex]

This must be equal to the LHS so we get.

[tex]

u \frac{dy}{dx}+y\frac{du}{dx}=u\frac{dy}{dx}+\frac{yu}{x+1}

[/tex]

For this to be true [tex]\frac{du}{dx}=\frac{u}{x+1} \Rightarrow u(x)=c(x+1)[/tex].

We can now write the differential equation as:

[tex]

\frac{duy}{dx}=x^2 u

[/tex]

You know u can you solve for y now?

Lets multiply the differential equation by u(x) this yields [itex]u(x)y'(x)+\frac{y(x)u(x)}{x+1}=u(x)x^2[/itex]. We want to write the left hand side as [itex]d(uy)/dx[/itex].

[tex]

\frac{duy}{dx}=u \frac{dy}{dx}+y\frac{du}{dx}

[/tex]

This must be equal to the LHS so we get.

[tex]

u \frac{dy}{dx}+y\frac{du}{dx}=u\frac{dy}{dx}+\frac{yu}{x+1}

[/tex]

For this to be true [tex]\frac{du}{dx}=\frac{u}{x+1} \Rightarrow u(x)=c(x+1)[/tex].

We can now write the differential equation as:

[tex]

\frac{duy}{dx}=x^2 u

[/tex]

You know u can you solve for y now?

Last edited:

- #5

CompuChip

Science Advisor

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While Cyosis is editing, you could try that last part?Sorry for asking a question? Anyway what do you know about integrating factors? Multiply the differential equation by u(x) first then try to write the LHS as d(uy)/dx.

- #6

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With you so far.You are pretty close. I am sure you can find a guide somewhere, but I have not much to do currently so I'll try to help you out.

Lets multiply the differential equation by u(x) this yields [itex]u(x)y'(x)+\frac{y(x)u(x)}{x+1}=u(x)x^2[/itex]. We want to write the left hand side as [itex]d(uy)/dx[/itex].

[tex]

\frac{duy}{dx}=u \frac{dy}{dx}+y\frac{du}{dx}

[/tex]

This must be equal to the LHS so we get.

[tex]

u \frac{dy}{dx}+y\frac{du}{dx}=u\frac{dy}{dx}+\frac{yu}{x+1}

[/tex]

For this to be true [tex]\frac{du}{dx}=\frac{u}{x+1} \Rightarrow u(x)=c(x+1)[/tex].

And now you've lost me.We can now write the differential equation as:

[tex]

\frac{duy}{dx}=x^2 u

[/tex]

You know u can you solve for y now?

*thinks hard*

Okay given that expression for [tex]\frac{duy}{dx}[/tex] I can substitute in the u(x), split the variables and integrate but I can't see how you arrived there in the first place.

In an ideal world I'd probably like to also know exactly what [tex]duy/dx[/tex] actually means, but just learning how to do these should be enough for now.

- #7

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Ohh... the penny has dropped.

Thankyou so much.

Edit: and I think I figured out what duy/dx means too!

Thankyou so much.

Edit: and I think I figured out what duy/dx means too!

Last edited:

- #8

Cyosis

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Hehe you're welcome.

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