Asked by paul
Please help. I have an implicit answer to a differential equation of 18y*sqrt(y)=(2(x^2-6x+23)^3/2)/3 plus arbitrary constant. What is the particular solution to this for which y=2 when x=1. And then express this particular solution answer in an explicit form. Please help and many thanks.
Answers
Answered by
drwls
Choose the value of the arbitary constsnt C so that y = 2 when x = 1.
That is all there is to it.
36sqrt2 = 2*[2*18^(3/2)]/3 + C
= 2*(4/3)27*sqrt2) + C
= 72 sqrt2 + C
C = -36 sqrt2
You can write the explicit solution for y(x) by taking the 2/3 root of y^(3/2)/18, with the C term inserted
That is all there is to it.
36sqrt2 = 2*[2*18^(3/2)]/3 + C
= 2*(4/3)27*sqrt2) + C
= 72 sqrt2 + C
C = -36 sqrt2
You can write the explicit solution for y(x) by taking the 2/3 root of y^(3/2)/18, with the C term inserted
Answered by
paul
Hello drwls. Please can you explain how you get 72sqrt2+c. Surely, if it is 50.911=50.911+c, if one subtracts either side c=0? Am I missing something?Thank you for your help.
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