# Solving differential equation involving initial value?

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)et

## Answers and Replies

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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)et

## The Attempt at a Solution

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
Homework Helper
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.

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

Mark44
Mentor
To elaborate on what Char. Limit said, separate the original equation like this:
$$\frac{dy}{y - 5} = dt$$
Now integrate both sides.

I took the integral and got

ln(y-5) = t+c

taking e of both sides gives

(y-5) = et+C

I do y(0) = y0 =(y-5) = et+C

How do I get my C term?

Dick
Homework Helper
I took the integral and got

ln(y-5) = t+c

taking e of both sides gives

(y-5) = et+C

I do y(0) = y0 =(y-5) = et+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?

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