Separation of Variables Substitution

Metronome

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Textbook problem: Consider the equation y' = f(at + by + c), where a, b, and c are constants. Show that the substitution x = at + by + c changes the equation to the separable equation x' = a + bf(x). Use this method to find the general solution of the equation y' = (y + t)^2.

Since nearly everything else given included general functions, I started by implicitly differentiating x = at + by + c, yielding dx = adt + bdy; with respect to t, it follows that x' = a + by', thus x' = a + bf(x), as the problem states. However, I don't understand what this achieves with the goal of somehow applying it to the second part of the question. I also don't understand how I might have discovered the given equation in order to make other, similar non-separable equations separable.
 
Textbook problem: Consider the equation y' = f(at + by + c), where a, b, and c are constants. Show that the substitution x = at + by + c changes the equation to the separable equation x' = a + bf(x). Use this method to find the general solution of the equation y' = (y + t)^2.

Since nearly everything else given included general functions, I started by implicitly differentiating x = at + by + c, yielding dx = adt + bdy; with respect to t, it follows that x' = a + by', thus x' = a + bf(x), as the problem states. However, I don't understand what this achieves with the goal of somehow applying it to the second part of the question. I also don't understand how I might have discovered the given equation in order to make other, similar non-separable equations separable.

I would write the ODE as:

\(\displaystyle \displaystyle \frac{dy}{dt}=f(at+by+c)\)

Now, given the substitution:

\(\displaystyle \displaystyle x(t)=at+by+c\)

We find:

\(\displaystyle \displaystyle \frac{dx}{dt}=a+b\frac{dy}{dt}\)

And so substituting from our original ODE, we may write:

\(\displaystyle \displaystyle \frac{dx}{dt}=a+bf(at+by+c)\)

Okay, so far so good. Next we are to consider:

\(\displaystyle \displaystyle \frac{dy}{dt}=(y+t)^2\)

Let's use the substitution:

\(\displaystyle \displaystyle x(t)=y+t\)

Hence:

\(\displaystyle \displaystyle \frac{dx}{dt}=\frac{dy}{dt}+1\)

And so the given ODE becomes:

\(\displaystyle \displaystyle \frac{dx}{dt}=x^2+1\)

Now we have a separable ODE that can be solved via integration. :)

We can apply the formula we derived in the first part of the problem, by observing \(\displaystyle f(x)=x^2\), and we have:

\(\displaystyle a=1,\,b=1,\,c=0\). :)
 
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So to reiterate at a higher level, first we proved that from dy/dt = f(at + by + c), the substitution x = at + by + c yields dx/dt = a + bf(x). The moral of the story I see here is just "u-substitution from calculus works in differential equations too!" Then we computed a concrete example, dy/dt = (y + t)^2. The substitution (let's use a different letter) u = y + t, where I assume u is decided via the same principles as with u-substitution from integral calculus, yields dx/dt = x^2 + 1, which can be solved and then substituted back to an equation in y's and t's. Does this sound correct?

As for the final sentence where you solve for the constants, is this somehow combining the two parts of the problem? Is it now f(u) = u^2, or are you referring to the x from the first part?
 
When I say we observe that \(\displaystyle f(x)=x^2\), I'm just saying that we see the output of the function is the square of the input.
 
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