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#2**+10 **

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} = e^{t}

So we have

e^{t} u'(t) + e^{t} u(t)= ke^{t}

And because of our employment of the integrating factor, we can write the left side in the form of the product rule, thusly....

[ e^{t} u(t) ] ' = ke^{t} ....integrate both sides

∫[ e^{t} u(t) ] ' dt = ∫ ke^{t} dt

e^{t} u(t) = ke^{t} + C divide both sides by e^{t}

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

---------------------------------------------------------------------------------------------------------------

CPhill Jan 1, 2015

#1**+10 **

u'(t) + u(t) = k

---> u'(t) = k - u(t)

---> ∫u'(t)dt = ∫[k - u(t)]dt

---> u(t) + C_{1} = ∫kdt - ∫u(t)dt

---> u(t) + C_{1} = kt + C_{2} - ∫u(t)dt

---> u(t) = kt - ∫u(t)dt + C

geno3141 Jan 1, 2015

#2**+10 **

Best Answer---------------------------------------------------------------------------------------------------------------

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} = e^{t}

So we have

e^{t} u'(t) + e^{t} u(t)= ke^{t}

And because of our employment of the integrating factor, we can write the left side in the form of the product rule, thusly....

[ e^{t} u(t) ] ' = ke^{t} ....integrate both sides

∫[ e^{t} u(t) ] ' dt = ∫ ke^{t} dt

e^{t} u(t) = ke^{t} + C divide both sides by e^{t}

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

CPhill Jan 1, 2015