Degree of diff eqn

(y''')^ (4/3) +(y')^ (1/5) + y = 0
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Sorry, I'm new here !!
The question is to find the degree of this diff eqn.

I have attached my working so far...The main problem is I don't know how to eliminate the fractional powers...

Pls tell me where I went wrong or what should be done further !
 

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The power of the highest order derivative after the eqn has been made integral in all of its derivatives
 
Very good! And the equation you are asking about is (y''')^(4/3)+ (y')^{1/5)+ y= 0.

What is the highest order derivative?
What is its power?

(Your text may require integer powers. To get rid of the fractional powers, you would have to take the 15th power of both sides. What is the power of the highest order derivative then? You do not need to actually do the entire multiplication to answer that.)
 
...To get rid of the fractional powers, you would have to take the 15th power of both sides. What is the power of the highest order derivative then? You do not need to actually do the entire multiplication to answer that.)

I don't think this would work. Please see the photo in post#3. Fractional powers will remain after raising to the 15th power.

I can't currently think of a way to solve this. Perhaps a substitution might be possible to remove the fractional powers, but I'm not sure what substitution would do the trick.
 
Fractional powers will remain but that is not relevant. The highest power of the highest derivative, y''', will be \(\displaystyle ((y''')^{4/3})^{15}= (y''')^{4(5)}= (y''')^{20}\).
 
Fractional powers will remain but that is not relevant. The highest power of the highest derivative, y''', will be \(\displaystyle ((y''')^{4/3})^{15}= (y''')^{4(5)}= (y''')^{20}\).
Hmmm, why not cube then, if the remaining powers don't matter? \(\displaystyle ((y''')^{4/3})^{3}= (y''')^{4}\).
 
(y''')^(4/3)+ (y')^{1/5)+ y= 0.

(y''')^(4/3) = -[ (y')^{1/5)+ y]

(y''')^(4) = -[ (y')^[3/5)+ y^3 + 3*y^2*(y')^(1/5) +3*y*(y')^(2/5)]

(y'''^4 + y^3)/(3*y) = -y'^(1/5) * [ (y')^(1/5)+y] ................................??? ....................... uncle

I give up ...... but I think HoI is correct
 
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