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# Worked example: separable differential equations

Two worked examples of finding general solutions to separable differential equations.

## Want to join the conversation?

• I don't understand what Sal did with the constant in the second example.
The anti-derivative of sin(x) is -cos(x) + C. After that he multiplies it by -1, and he gets cos(x) + C. Shouldn't that be cos(x) - C?
• Remember, 𝐶 = ℝ. That is, the constant of integration can be any real number. Writing +𝐶 is the same as writing -𝐶 because 𝐶 encapsulates all real numbers. That is, if we fix 𝐶₁ as one possible value of 𝐶, then since {𝐶₁ ∪ -𝐶₁} ∈ ℝ = 𝐶, the sign of the constant of integration does not matter.

For example, an antiderivative of sin 𝑥 is -cos 𝑥 + 5. Multiplying this by -1 for whatever reason gives us cos 𝑥 – 5. However, another antiderivative of sin 𝑥 is -cos 𝑥 – 5. Multiplying this by -1 gives us cos 𝑥 + 5. So cos 𝑥 ± 5 are both valid results from multiplying antiderivatives of sin 𝑥 by -1. This same logic applies to all real numbers (𝐶) so whether or not we distribute the negative to 𝐶 is irrelevant. It still encapsulates ℝ no matter what its sign is.

Comment if you have questions!
• Would it be fine if I simplify my answer to y = sec(x) + C?
• Tempting but no. That would only work if the solution were:
y = (1 / cos x) + C
But the answer was
y = 1 / (cos x + C)
and you can't pull the Constant out of the denominator.
• At 1 :48 why Sal is putting c on right hand side only but not with y^2/2. Please explain.
• Greetings!

A few videos back, Sal mentioned that if you put C on both sides, then you subtract C from both sides, you simply end up with C on the right side. So, it is a shortcut to simply leave the C off of the Y (left side) of the equation as once you solve for Y, you will only end up with +C on the right side anyway.
• I didn't get the idea behind the "general solution" and "Particular solution"
Could someone help me please
• that + C at the end after integrating casues it to be a general solution, since we don't know what the exact solution is. Or maybe another way to think of it is that C can be any number, and if you differentiated it you'd get the exact same answer. think x^2+1, x^2+2 and even just x^2. They all differentiate to 2x.

It isn't until we have some initial conditions we can figure out what that C is. initial conditions need to be when x is something then y is something else. Then you can plug these into the equation once you have things in terms of y and use algebra to figure out what C is. Does that help?
• That was strangely fun Sal, im ecstatic over here
• yeah, calc is fun :D
• I've heard people say "At , it was A-OK for Sal to keep C as C when multiplying by -1, because C is still just an arbitrary constant."

That I understand.

This I do not understand:

At , Sal takes the reciprocal of both sides. And he puts the C in the denominator. But if you have and arbitrary constant C and you raise it to the -1 power, it'll still be an arbitrary constant!! It's domain hasn't changed.
• We cannot take the reciprocal of C if C=0. Keeping C in the denominator reflects that.
• Isn't y = 0 also a solution for the 2nd one? It's not given by the general solution he gave.
• It is a solution, and it's not given because Sal started by dividing both sides by y. Therefore, he implicitly assumed that y≠0, and missed that solution. After solving, Sal should've gone back and checked the y=0 case.
• In the second example at time can I integrate the y side into ln|y²| * 1/2 ?
• That is the integral of 1/y (since by logarithm properties, it's just ln|y|), but why would you integrate again? You already have y isolated.