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5 votes
Is y=x^3 a solution of the differential equation yy'=x^5+y

User Skarist
by
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1 Answer

3 votes

No; we have
y=x^3\implies y'=3x^2. Substituting these into the DE gives


3x^5=x^5+x^3

which reduces to
x^3=0, true only for
x=0.

User Amol Brid
by
7.1k points
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