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

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SUMMARY

The solution to the initial value problem defined by the differential equation dy/dx = y - 3 with the initial condition y(0) = 4 is derived through separation of variables. The equation is transformed into the form 1/(y-3) dy = dx, leading to the integral ln(y-3) = x + C. By exponentiating both sides, the general solution is expressed as y = Ce^x + 3. Applying the initial condition allows for the determination of the constant C, resulting in the specific solution.

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

y-3 = Ce^x
y = Ce^x + 3
 
Mitchtwitchita said:
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.
 

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