AP®︎/College Calculus AB
Course: AP®︎/College Calculus AB > Unit 7Lesson 6: Finding particular solutions using initial conditions and separation of variables
- Particular solutions to differential equations: rational function
- Particular solutions to differential equations: exponential function
- Particular solutions to differential equations
- Worked example: finding a specific solution to a separable equation
- Worked example: separable equation with an implicit solution
- Particular solutions to separable differential equations
Worked example: separable equation with an implicit solution
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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- At2:25Sal 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.(7 votes)
- 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.(16 votes)
- What on earth is an explicit function?(8 votes)
- Any equation that can be written as y = f(x). Implicit equations like x^5y + y^2 - 1 = y^3 cannot.(11 votes)
- 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?(2 votes)
- 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.(8 votes)
- 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?(2 votes)
- We can't. The inverse of sin(y)+2y cannot be expressed in terms of any of the familiar functions.(2 votes)
- 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.(2 votes)
- 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.(2 votes)
- How do we know when it's (cos(y+2)) or (cos(y)+2)? Sal did this at1:06.(1 vote)
- [Teacher] We're given a differential equation right over here, cosine of y plus two, this whole thing times the derivative of y with respect to x is equal to two x, and we're given that for a particular solution, when x is equal to one, y of one is equal to zero, and we're asked what is x when y is equal to pi? So the first thing I like to look at when I see a differential equation is: is it separable? Can I get all the ys and dys on one side, and can I get all the xs and dxs on the other side? And this one seems like it is. If I multiply both sides by dx, where you can view dx as the x differential of an infinitely small change in x, well, then you get cosine of y plus two times dy is equal to two x times dx, so just like that I would've been able to, all I did is I multiplied both sides of this times dx, and I was able to separate the ys and the dys from the xs and the dxs, and now I can integrate both sides. So, if I integrate both sides, what am I going to get? So the antiderivative of cosine of y with respect to y, with respect to y, is sine of y, and then the antiderivative of 2 with respect to y is 2y, and that is going to be equal to, well, the antiderivative of 2x with respect to x is x squared, and we can't forget that we could say a plus a different constant on either side, but it serves our purpose just to say plus c on one side, and so this is a general solution to the separable differential equation, and then we can find the particular one by substituting in when x is equal to one, y is equal to zero, so let's do that to solve for c. So we get, or when y is equal to zero, x is equal to one, so, sine of zero plus two times zero, all I did is I substituted in the zero for y, is equal to x squared, well now x is one, is equal to one squared plus c. Well sine of zero is zero, two times zero is zero, all of that's just gonna be zero, so we get zero is equal to one plus c, or c is equal to negative one. So now we can write down the particular solution to this differential equation that meets these conditions. So we get, let me write it over here, sine of y plus two y is equal to x squared, and our constant is negative one, so minus one. And now what is x when y is equal to pi? So, sine of pi plus two times pi is equal to x squared minus one, see sine of pi is equal to zero, and so we get, let's see we can add one to both sides, and we get two pi plus one is equal to x squared, or we could say that x is equal to the plus or minus the square root of two pi plus one. So I would write the plus or minus square root of two pi plus one, and we're done.