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

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{4y}{x^2}\implies\dfrac{\mathrm dy}y=\dfrac4{x^2}\,\mathrm dx$

Integrate both sides to get

$\ln|y|=-\dfrac4x+C$

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

$\ln|e|=-\dfrac4{-4}+C\implies C=0$

so the particular solution is

$\ln|y|=-\dfrac4x\iff y=e^{-4/x}$

Answer 2

y = e^(-4/x)

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i took the test so i know this is the answer

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