u'(t)+u(t)=k ...this is a linear differential equation which is in the form du/dt + p(t) u = g(t)......note that p(t) = 1
We need to multiply through by an integrating factor which is given by e∫p(t)dt = e∫(1)dt = et
So we have
et u'(t) + et u(t)= ket
And because of our employment of the integrating factor, we can write the left side in the form of the product rule, thusly....
[ et u(t) ] ' = ket ....integrate both sides
∫[ et u(t) ] ' dt = ∫ ket dt
et u(t) = ket + C divide both sides by et
u(t) = k + Ce-t
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Note that this solution "works" because
u'(t) = -Ce-t and u(t) = k + Ce-t
And their sum = k
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u'(t) + u(t) = k
---> u'(t) = k - u(t)
---> ∫u'(t)dt = ∫[k - u(t)]dt
---> u(t) + C1 = ∫kdt - ∫u(t)dt
---> u(t) + C1 = kt + C2 - ∫u(t)dt
---> u(t) = kt - ∫u(t)dt + C
u'(t)+u(t)=k ...this is a linear differential equation which is in the form du/dt + p(t) u = g(t)......note that p(t) = 1
We need to multiply through by an integrating factor which is given by e∫p(t)dt = e∫(1)dt = et
So we have
et u'(t) + et u(t)= ket
And because of our employment of the integrating factor, we can write the left side in the form of the product rule, thusly....
[ et u(t) ] ' = ket ....integrate both sides
∫[ et u(t) ] ' dt = ∫ ket dt
et u(t) = ket + C divide both sides by et
u(t) = k + Ce-t
---------------------------------------------------------------------------------------------------------------
Note that this solution "works" because
u'(t) = -Ce-t and u(t) = k + Ce-t
And their sum = k
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