Hello, I would greatly appreciate any help that I can have with the following calculus question below. It has to do with separable differential equations. I'm also going to explain what I've done so far, so please bear with me (but honestly I don't think I've done much).
Consider this: dy/dx = (y-1)/(x^2)
The Question: Find the Particular Solution y=f(x) with the initial condition f(1)=2
If you don't want to look through all of my work, in short, I just need help solving for y in this equation:
ln|y-1| = ln|x^2|
solving this for y is where I'm stuck :(
If you'd like to see how I got this equation, all of my work is below.
What I have done so far: First I tried getting "dy" and the "y's" on the same side of the equal sign and the "dx" and the "x's" on the same side of the equal sign.
I did this by multiplying and dividing like so:
(multiply "dx" on both sides)
(dx) dy/dx = (y-1)/(x^2) (dx)
(simplify)
dy= (1 / (x^2)) (y-1) dx
(divide "y-1" on both sides)
(dy) / (y-1) = (1 / (x^2) dx) / (y-1)
(simplify)
1 / (y-1) dy = 1/(x^2) dx
Second, I integrated both sides using a definite integral, simplified what I got after integrating, and I got this:
ln|y-1| = ln|x^2| + c
Third, I considered how to question mentioned that f(1)=2, so I substitute "1" for "x" and "2" for y and got this:
(substitute)
ln|(2)-1| = ln|(1)^2| + c
(Simplify)
ln|1| = ln|1| + c
I then solved for "c", and found that c = 0
I substitute "0" for "c" in the equation I got after integrating before (the one from my second step) and got this:
ln|y-1| = ln|x^2|
If everything I did was correct, I just have to solve the above equation for y, BUT I don't know how :(
dy/dx = (y-1)/(x^2)
dy = (y-1)/(x^2) dx
dy/(y-1) = x^-2 dx
the next step is where you went wrong,
ln (y-1) = (-1/3)x^-3 + c
check by taking the derivative.
given: the point (1,2) lies on it, so
ln (2-1) = (-1/3)(1^3) + c
0 = -1/3 + c
c = 1/3
ln (y-1) = (-1/3)x^-3 + 1/3
= ( -1/x^3 + 1)/3
= (1 - 1/x^3)/3
ln (y-1) = ( x^3 -1)/(3x^3)
which means:
y - 1 = e^( ( x^3 -1)/(3x^3) )
y = e^( ( x^3 -1)/(3x^3) ) + 1
check this, haven't done a lot of these in 60 years.
Thank you very much, you opened my eyes to my error, however I see a slight mistake in your work as well. You got here:
dy/dx = (y-1)/(x^2)
dy = (y-1)/(x^2) dx
dy/(y-1) = x^-2 dx
this is all correct, but when you integrate it should be this:
ln (y-1) = -1/x + c
(because the antiderivative of x^-2 is -1/x +c)
then I work out all of everything afterward, I find that c = 1
I substitute 1 for c and get this equation
ln|y-1| = -1/x +(1)
But now I am back at my original problem.
I have to solve for y, but don't know how :(
thanks for catching that, of course when you add 1 to -2 you get -1, not -3
- sometimes the simple things mess us up!
Ok, then I should have had:
ln (y-1) = -1/x + c
and plugging in (1,2)
ln 1 = (-1/1) = c
c = 1, you had that.
so we get to
ln (y-1) = -1/x + 1 or (x-1)/x
now recall you we change logs to exponentials
e.g. log 2 8 = 3 <-----> 2^3 = 8
ln 5 = x <------> e^x = 5
etc
so ....
ln (y-1) = (x-1)/x
means
y - 1 = e^( (x-1)/x )
y = e^( (x-1)/x ) + 1
what do you think?
OK. Now you have
ln(y-1) = -1/x + 1
y-1 = e^(-1/x + 1)
y = e^(1 - 1/x) + 1
I think we have a golden answer!
Thank you so much for your wonderful help, it means a lot that you'd take the time to help me out and teach me something along the way.
I have no idea how I made it to a calculus class without knowing how to solve logarithms, and yet, here I am.
Thanks Again!
Thank you too steve!
To solve the equation ln|y-1| = ln|x^2|, we can use the property of logarithms which states that ln(a) = ln(b) if and only if a = b. In this case, we can apply this property to simplify the equation.
ln|y-1| = ln|x^2|
Since the natural logarithm function is strictly increasing for positive arguments, we can drop the absolute value signs:
y - 1 = x^2
Now, we can isolate y by adding 1 to both sides of the equation:
y = x^2 + 1
This is the particular solution y = f(x) to the differential equation dy/dx = (y-1)/(x^2) with the initial condition f(1) = 2.