Determine which function is a solution to the differential equation xy' + 2y = 0.
e^2x, x², x^-2, None of these

Solution:

Given, xy’ + 2y = 0

We know, y’ = dy/dx

So, the differential equation becomes

\(x\frac{dy}{dx}+2y=0\)

Now, \(x\frac{dy}{dx}=-2y\)

\(-\frac{1}{2y}dy=\frac{1}{x}dx\)

Integrating both sides,

\(\\\int -\frac{1}{2y}dy=\int \frac{1}{x}dx\\-\frac{1}{2}\int \frac{1}{y}dy=\int \frac{1}{x}dx\)

\(-\frac{1}{2}ln\, y=ln\, x+C\)

On rearranging,

ln y = -2 ln x + C

\(y=e^{-2\, ln\, x+C}\)

\(y=e^{-2\, ln\, x}.e^{C}\)

\(y=Ce^{ln\, (x^{-2})}\)

\(y=Cx^{-2}\)

\(y=\frac{C}{x^{2}}\)

Therefore, \(y=\frac{C}{x^{2}}\)

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Determine which function is a solution to the differential equation xy' + 2y = 0.

Summary:

The solution to the differential equation xy' + 2y = 0 is \(x^{-2}\).