First-Order Differential Equation

[sandy.bridge]

Homework Statement

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?

 

[Dick]

No, no. Your last line should read,

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

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?

 

[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?

 

[SammyS]

Homework Statement

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


...

Why not integrate the right hand side ?

 

[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!

 

[Dick]

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

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?

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

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.