Question

Find the specific solution of the differential equation dy dx equals the quotient of 4 times y and x squared with condition y(-4) =

e.

Answer 1

y = e^(-4/x)

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i took the test so i know this is the answer

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Answer 2

`(\mathrm dy)/(\mathrm dx)=(4y)/(x^2)\implies\frac{\mathrm dy}y=\frac4{x^2}\,\mathrm dx`

Integrate both sides to get


`\ln|y|=-\frac4x+C`

Given that
`y(-4)=e`, we have


`\ln|e|=-\frac4{-4}+C\implies C=0`

so the particular solution is


`\ln|y|=-\frac4x\iff y=e^(-4/x)`