First-Order Differential Equation

  • #1
sandy.bridge
798
1

Homework Statement




Find x(t) given the following:
x(0)=2
[tex]dx/dt+4x=10e^{-2t}[/tex]
[tex]e^{4t}(dx/dt+4x)=10e^{-2t}e^{4t}[/tex]
[tex]d/dt(e^{4t}x)=10e^{2t}[/tex]
[tex]\int{d(e^{4t}x)}=\int{10e^{2t}dt}[/tex]
[tex]e^{4t}x+C=\int{10e^{2t}}[/tex]
[tex]e^{4t}x=\int{10e^{2t}}+C_1[/tex]
[tex]x(t)=\int{10e^{-2t}}+e^{-4t}C_1[/tex]
From here I get stuck. I know I can integrate it, but then I have another constant?
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,263
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No, no. Your last line should read,

[tex]x(t)=e^{-4t}\int{10e^{2t}}+e^{-4t}C_1[/tex]

You can't just move the e^(-4t) inside the integral like that. Now sure, integrate it. If you add another constant, does it matter?
 
  • #3
sandy.bridge
798
1
[tex]x(t)=e^{-4t}\int{10e^{2t}}+e^{-4t}C_1[/tex]
[tex]x(t)=e^{-4t}(5e^{2t}+C_2)+e^{-4t}C_1[/tex]

Then, since
[tex]x(0)=2[/tex]
we have
[tex]-3=C_2+C_1[/tex]
Now if I combine these constant to make a new one, that doesn't really help me out in regards to the equation does it?
 
  • #4
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
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1,189

Homework Statement




Find x(t) given the following:
x(0)=2
[tex]dx/dt+4x=10e^{-2t}[/tex]
[tex]e^{4t}(dx/dt+4x)=10e^{-2t}e^{4t}[/tex]
[tex]d/dt(e^{4t}x)=10e^{2t}[/tex]
[tex]\int{d(e^{4t}x)}=\int{10e^{2t}dt}[/tex]
[tex]e^{4t}x+C=\int{10e^{2t}}dt[/tex]


...
Why not integrate the right hand side ?
 
  • #5
sandy.bridge
798
1
@ Sammy:
I was doing that before, but I could not seem to get the answer. I just tried again given the information Dick presented and got the right answer. Thanks!
 
  • #6
Dick
Science Advisor
Homework Helper
26,263
620
[tex]x(t)=e^{-4t}\int{10e^{2t}}+e^{-4t}C_1[/tex]
[tex]x(t)=e^{-4t}(5e^{2t}+C_2)+e^{-4t}C_1[/tex]

Then, since
[tex]x(0)=2[/tex]
we have
[tex]-3=C_2+C_1[/tex]
Now if I combine these constant to make a new one, that doesn't really help me out in regards to the equation does it?

[tex]x(t)=e^{-4t}(5e^{2t}+C_2)+e^{-4t}C_1[/tex]

is the same as

[tex]x(t)=5e^{-2t}+e^{-4t}(C_2+C_1)[/tex]

You may as well just call C=C1+C2. The extra constant doesn't do anything different than the first one did.
 

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