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Hello, could you advise me how to solve this please? ==> u'(t)+u(t)=k

 Jan 1, 2015

Best Answer 

 #2
avatar+98130 
+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 = 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

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

 

 Jan 1, 2015
 #1
avatar+17746 
+10

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

 Jan 1, 2015
 #2
avatar+98130 
+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 = 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

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

 

CPhill Jan 1, 2015
 #3
avatar+99315 
+5

Great answers from both of you :)

I love your new icon Chris     

 Jan 2, 2015

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