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AP®︎/College Calculus AB

Course: AP®︎/College Calculus AB>Unit 7

Lesson 6: Finding particular solutions using initial conditions and separation of variables

Worked example: separable equation with an implicit solution

AP.CALC:
FUN‑7 (EU)
,
FUN‑7.E (LO)
,
FUN‑7.E.1 (EK)
,
FUN‑7.E.2 (EK)
,
FUN‑7.E.3 (EK)
Sometimes the solution of a separable differential equation can't be written as an explicit function. This doesn't mean we can't use it!

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• At Sal says that C is -1. But if we would add plus c to the other side of the equation so that we have sin(y)+2y + c = x^2 it results in c= 1 not -1. Why is -1 right if it is? Is there a rule a to analyse where to add c. The rest is clear to me only that constant adding looks strange.
• If you think about it, moving C from one side to the other will require you to add or subtract it depending on its sign. Adding -1 to the other side will result in 1 vice versa subtracting the 1 to other side will result in -1. Both answers are correct.
• What on earth is an explicit function?
• Any equation that can be written as y = f(x). Implicit equations like x^5y + y^2 - 1 = y^3 cannot.
• So working through lots of problems from calculus and physics and such I hear them mention anti-derivative. I feel like this is just integration but some videos use integrate and others anti-derivative and they dont switch back and forth in the same video. Is this just integration?
• An integral is the area under a curve, or a function that gives the area under a curve in terms of the length of the interval. An antiderivative of f is a function whose derivative is f.

The two quantities are defined very differently, but they happen to be mathematically equivalent. This is the statement of the fundamental theorem of calculus.
• this equation is separable, still it seems that we can't find a particular solution for y=f(x). Here, we got siny + 2y = x^2-1. how can we now find y?
• We can't. The inverse of sin(y)+2y cannot be expressed in terms of any of the familiar functions.
• The question tells us that "y(1) = 0"...so I think this means either "(cos(1)+2) dy/dx = 2x = 0" (the original problem) or "sin(1)+2(1) = x^2 + C = 0" (the separated problem). But Sal goes on to solve "sin(0)+2(0) = 1^2 + C". Did Sal get mixed up or am I not understanding something?
(1 vote)
• You're misreading what y(1) = 0 means.
y(1) = 0 should be read that "at a value of x=1, y is equal to 0".
You can also think of y(1) = 0 as the point (x,y)=(1,0).
You were substituting the point (0,1) instead of (1,0) when you were solving for C. Sal was correct.
• Find the particular solution to the differential equation dy/dx=xy+xwhich satisfies y = 3 when x = 0.
(1 vote)
• Why didn't we have y alone on the left side before using the initial value y(1)=0 when trying to find out the value of the constant c? Because that's how it was done in previous videos/lessons, and when I do it like that I get that c=- (1/2)
(1 vote)
• That's what 'implicit solution' means – we don't have 𝑦 as a function of 𝑥, instead we have sin 𝑦 + 2𝑦 as a function of 𝑥.

If you like you can go to desmos.com and type in the implicit solution to get a graph of the explicit solution.

By trying different values for 𝐶, you'll see that 𝐶 = −1 is the only value that satisfies the initial condition.