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==Separable (homogeneous) first-order linear ordinary differential equations==
{{see also|Separable partial differential equation}}
A separable ''linear'' [[ordinary differential equation]] of the first order must be homogeneous and has the general form
must be homogeneous and has the general form
:<math>\frac{dy}{dt} + f(t) y = 0</math>
where <{{math>|''f'' (''t'')</math>}} is some known [[function (mathematics)|function]]. We may solve this by [[separation of variables]] (moving the ''{{mvar|y''}} terms to one side and the ''{{mvar|t''}} terms to the other side),
:<math>\frac{dy}{y} = -f(t)\, dt</math>
Since the separation of variables in this case involves dividing by ''{{mvar|y''}}, we must check if the constant function {{math|1=''y=0'' = 0}} is a solution of the original equation. Trivially, if {{math|1=''y=0'' = 0}} then {{math|1=''y''′ = 0''}}, so {{math|1=''y=0'' = 0}} is actually a solution of the original equation. We note that {{math|1=''y=0'' = 0}} is not allowed in the transformed equation.
We solve the transformed equation with the variables already separated by [[Integral Calculus|Integratingintegrating]],
:<math>\ln |y| = \left(-\int f(t)\,dt\right) + C</math>
where ''{{mvar|C''}} is an arbitrary constant. Then, by [[exponentiation]], we obtain
:<math>y = \pm e^{\left(-\int f(t)\,dt\right) + C} = \pm e^{C} e^{-\int f(t)\,dt}.</math>.
Here, <{{math>|''e^{C}''<sup>0''C''</mathsup> > 0}}, so <math>\pm e^{C} \neq 0</math>. But we have independently checked that {{math|1=''y=0'' = 0}} is also a solution of the original equation, thus
:<math>y = A e^{-\int f(t)\,dt}.</math>.
with an arbitrary constant ''{{mvar|A''}}, which covers all the cases. It is easy to confirm that this is a solution by plugging it into the original differential equation:
:<math>\frac{dy}{dt} + f(t) y = -f(t) \cdot A e^{-\int f(t)\,dt} + f(t) \cdot A e^{-\int f(t)\,dt} = 0</math>
Some elaboration is needed because {{math|''f''&fnofthinsp;''(''t'')}} mightmay not even be integrable. One must also assume something about the domains of the functions involved before the equation is fully defined. The solution above assumes the [[real number|real]] case.
If <{{math>|1=''f''(''t'') =\alpha</math> ''α''}} is a constant, the solution is particularly simple, <math>y = A e^{-\alpha t}</math> and describes, e.g., if <{{math|''α'' >\alpha> 0</math>}}, the exponential decay of radioactive material at the macroscopic level. If the value of <math>\alpha</math>{{mvar|α}} is not known a priori, it can be determined from two measurements of the solution. For example,
:<math>\frac{dy}{dt} + \alpha y = 0, y(1)=2, y(2)=1</math>
gives <{{math>\alpha|1=''α'' = \ln(2)</math>}} and <math>y = 4 e^{-\ln(2) t}= 2^{2-t}</math>.
==Non-separable (non-homogeneous) first-order linear ordinary differential equations==
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