Solving differential equation involving initial value?

[Chandasouk]

dy/dt = y-5 , y(0) = y₀

The general form for first ODE has to resemble this

dy/dt + p(t) = g(t)

so I moved the y over to the left side of the equation

dy/dt - y = -5

I think this is where I screw things up. It's not really in this form dy/dt + p(t) = g(t) but rather dy/dt - p(t) = g(t).

Regardless, I went to find an integrating factor $$\mu$$(t) = e^{$$\int$$p(t)dt}

which gave me e^-t as my $$\mu$$(t). You then have to multiply that to both sides of this equation

dy/dt - y = -5

Is my procedure correct so far? Because I cannot get the correct answer which is

y=5+(y₀-5)e^t

 

[mathstew]

dy/dt = y-5 , y(0) = y₀

The general form for first ODE has to resemble this

dy/dt + p(t) = g(t)

so I moved the y over to the left side of the equation

dy/dt - y = -5

I think this is where I screw things up. It's not really in this form dy/dt + p(t) = g(t) but rather dy/dt - p(t) = g(t).

Regardless, I went to find an integrating factor $$\mu$$(t) = e^{$$\int$$p(t)dt}


which gave me e^-t as my $$\mu$$(t). You then have to multiply that to both sides of this equation

dy/dt - y = -5

Is my procedure correct so far? Because I cannot get the correct answer which is

y=5+(y₀-5)e^t

I'm not sure you wrote the problem correctly. You have dy/dt - y = -5 but y is not of the form p(t).

I get y=$$\frac{5t+y(0)}{1-t}$$

 

[Dick]

I'm not sure you wrote the problem correctly. You have dy/dt - y = -5 but y is not of the form p(t).

I get y=$$\frac{5t+y(0)}{1-t}$$

Well, that's wrong. The solution is an exponential. You can probably get it with an integrating factor. But it's a lot easier if you just realize the equation is separable.

 

[Chandasouk]

I actually am behind two lectures because I switched into a new math section, so could you show how to do it with integrating factor?

 

[Char. Limit]

There's no need. Just take the initial and divide both sides by (y-5). Then you have something integrable.

 

[Mark44]

To elaborate on what Char. Limit said, separate the original equation like this:
$$\frac{dy}{y - 5} = dt$$
Now integrate both sides.

 

[Chandasouk]

I took the integral and got

ln(y-5) = t+c

taking e of both sides gives

(y-5) = e^t+C

I do y(0) = y₀ =(y-5) = e^t+C

How do I get my C term?

 

[Dick]

I took the integral and got

ln(y-5) = t+c

taking e of both sides gives

(y-5) = e^t+C

I do y(0) = y₀ =(y-5) = e^t+C

How do I get my C term?

You can write e^(t+C) as C*e^t since C is an arbitrary constant. So put t=0 and y=y0 into y-5=C*e^t. What's C?

 

[Chandasouk]

Thanks, Dick. I wasn't aware that you could rewrite it that way. I'll have to keep that in mind. C is then y₀-5

 

[Dick]

Thanks, Dick. I wasn't aware that you could rewrite it that way. I'll have to keep that in mind. C is then y₀-5

e^(x+C)=e^C*e^x. Since C is arbitrary you may as well just call e^C the arbitrary constant and write it C*e^x.