Solution to Initial Value Problem: dy/dx = y-3, y(0)=4

[Mitchtwitchita]

The solution of the initial value problem, dy/dx = y-3. y(0) = 4, what does y=?

dy/y-3 = dx
1/y-3 dy = dx
ln(y-3) = ? + C

I have no idea how to do this one, can anybody please help me out? I'm having a hard time seeing how to separate the y and the -3 or if they are separable at all.

 

[sutupidmath]

The solution of the initial value problem, dy/dx = y-3. y(0) = 4, what does y=?

dy/y-3 = dx
1/y-3 dy = dx
ln(y-3) = ? + C

I have no idea how to do this one, can anybody please help me out? I'm having a hard time seeing how to separate the y and the -3 or if they are separable at all.

well you started it right:

$$\frac{dy}{dx}=y-3=>\frac{dy}{y-3}=dx$$

Now to get our answer we integrate both parts

$$\int\frac{dy}{y-3}=\int dx=>ln(y-3)=x+C$$

Now do you know how to isolate y by itself. Hint: remember that ln and the exponential function with base "e" are inverses. So, exponentiate both sides, and try to go as further as you can. If you're still stuck, ask again.

 

[Mitchtwitchita]

I do. thanks stupidmath. I guess it was the "invisible" x that got me.

y-3 = Ce^x
y = Ce^x + 3

 

[Integral]

I do. thanks stupidmath. I guess it was the "invisible" x that got me.

y-3 = Ce^x
y = Ce^x + 3

Now apply your initial conditons to find C.