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What is a solution to the differential equation #dy/dx=x-y#?

Answer 1

Charles Arocha

# y = x - 1 + C/e^x#

#dy/dx=x-y#

not separable, not exact, so set it up for an integrating factor

#dy/dx + y =x#

the IF is #e^(int dx) = e^x# so

#e^x dy/dx + e^x y =xe^x#

#d/dx (e^x y) =xe^x#

#e^x y = int xe^x \ dx qquad triangle#

for the integration, we use IBP: #int u v' = uv - int u' v#

#u = x, u' = 1# #v' = e^x, v = e^x#

#implies x e^x - int e^x \ dx#

#= x e^x - e^x + C#

so going back to #triangle#

#e^x y = x e^x - e^x + C#

# y = x - 1 + C/e^x#

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Answer 2

Axel Alvarez

The solution to the differential equation (\frac{dy}{dx} = x - y) is given by the function (y(x) = ce^{-x} + x - 1), where (c) is an arbitrary constant.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.