# First-Order Differential Equation

1. Jan 10, 2012

### sandy.bridge

1. The problem statement, all variables and given/known data

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

2. Jan 10, 2012

### Dick

$$x(t)=e^{-4t}\int{10e^{2t}}+e^{-4t}C_1$$

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. Jan 10, 2012

### sandy.bridge

$$x(t)=e^{-4t}\int{10e^{2t}}+e^{-4t}C_1$$
$$x(t)=e^{-4t}(5e^{2t}+C_2)+e^{-4t}C_1$$

Then, since
$$x(0)=2$$
we have
$$-3=C_2+C_1$$
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. Jan 10, 2012

### SammyS

Staff Emeritus
Why not integrate the right hand side ?

5. Jan 10, 2012

### sandy.bridge

@ 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. Jan 10, 2012

### Dick

$$x(t)=e^{-4t}(5e^{2t}+C_2)+e^{-4t}C_1$$

is the same as

$$x(t)=5e^{-2t}+e^{-4t}(C_2+C_1)$$

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