- #1

josephsuk

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**1. y''y^4 = 8**

**I tried almost every method I know, including laplace transforms, variation of parameters, reductin of order, v=y' substitution**

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- Thread starter josephsuk
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- #1

josephsuk

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- #2

jackmell

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First, you realize that y'' can be written as d(y') and y' as dy right? So what happens if you multiply both sides of the equation by y'?

Well, you get:

[tex]y'd(y')y^4=8dy[/tex]

Now what?

Well, you get:

[tex]y'd(y')y^4=8dy[/tex]

Now what?

Last edited:

- #3

josephsuk

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It appears to be logical to use separation of variables to get y'd(y')=8y^-4dy

and then integrate to get the solution?

and then integrate to get the solution?

Last edited:

- #4

jackmell

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It appears to be logical to use separation of variables to get [tex]y'd(y')=8 dy[/tex]

Forgot the other part. I get:

[tex]y'd(y')=8y^{-4} dy[/tex]

- #5

josephsuk

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Would integrating y' d(y') give me ((y')^2)/2 or something else?

If it does give me that answer, then I get y to be [tex]y=\frac{5}{2}x\sqrt[5]{\frac{-16}{3}}[/tex]

- #6

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Would integrating y' d(y') give me ((y')^2)/2 or something else?

If it does give me that answer, then I get y to be [tex]y=\frac{5}{2}x\sqrt[5]{\frac{-16}{3}}[/tex]

Look at your given DE and see if you think a constant times x can be a solution.

- #7

jackmell

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Would integrating y' d(y') give me ((y')^2)/2 or something else?

If it does give me that answer, then I get y to be [tex]y=\frac{5}{2}x\sqrt[5]{\frac{-16}{3}}[/tex]

No way dude. you got:

[tex]\int y' d(y')=\int 8 y^{-4} dy[/tex]

and that's

[tex]\frac{(y')^2}{2}=-8/3 y^{-3}+c[/tex]

Ok, can you now separate variables (take square root first), and then post what the next integral expression would be? Can't integrate it directly (not easily) but just the expression of what has to be integrated is good enough for now.

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