separable differential equation with an initial condition

Learn how to solve a separable differential equation with an initial condition. This is usually the first kind of differential equations that we learn in an ordinary differential equations class.

To learn how to solve different types of differential equations:

For more calculus & differential equation tutorials, check out @justcalculus

#DifferentialEquations

[sammo]

pretty sure the background music boosted my math skills by at least 30%

[NotYourAverageNothing]

C can be -1 as well as 1.

Reverse the process to check:

y = [sin(x) – 1]^2 – 1 Still, y(π) = 0. Keep going (although unnecessary).

y = sin^2(x) – 2sin(x)

dy/dx = 2sin(x)cos(x) – 2cos(x)

1/2 dy/dx = sin(x)cos(x) – cos(x)

1/2 dy/dx = [sin(x) – 1]*cos(x) From the top, *√*(y+1) = sin(x) – 1

1/2 dy/dx = *√*(y+1) cos(x)

[mohammedhajomar]

thanks a lot, your smiles makes me happy

[infinitymfg]

Oh no! You have to change your name to blackpenredpenbluepen!

[Markusepoiss]

But couldn't you just label C² to be another constant C2 and then you solve for that?

[Cannongabang]

bello blackpenredpen i got a question: when u solve integrals, You usually don't care whether what you write is sensed or not. If though we are operating in real, do we take in consideration the fact that when we operate, if the functions are not well defined, the integral doesn't exist ?
For example, when you write (second to last video): by arctan(k) i mean inverse tangent of k} the solution

arctan(y) = ½x² + C
it should follow
½x² + C included in
]-½π; ½π[
also denoted as
½π>½x² + C> -½π (&)
inf of x² is 0 so it comes pretty natural that:
½π > 1/2x² +C ≥ 0+C > -½π
so we have the condition(#):
½π > C > -½π
also, the first side of the disequation (&) means:
½π>½x² + C
½π - C > ½x²
½x² < ½π +(-C)
-C has a superior limit, ½π, by the condition(#)
½x² < ½π -C <π/2 + π/2 =π
½x²< π => x² < 2π
means the condition
|x| < √(2π) also known as
-√(2π) < x < √(2π)
This will limit both x and C);
and when you tangent both sides you get y = tg(½x² + C)

or in this video, when you write
√y+1 = sin(x) + C
y ≥-1 is probably trivial since you got that function from the beginning;
but √y+1≥ 0 for every real at the very least means the other side is positive aswell
sin(x) + C ≥ 0, so
sin(x) ≥ -C for every x;
inf of the values of sin(x) is -1 so we have
sin(x) ≥-1 ≥ -C
and the condition -1≥-C => C≥1;

here is the other question: are the operations you are doing so general that they work as a general method for solving integrals in the complex set (so, all the conditions i am setting are just useless)
or.. is it that
since the solution is
√y+1 = sin(x) + C
y= (sinx +C)² -1 no matter what the C will just work fine for our derivation?

[Ahinator]

y= Sin(x)*(sin(x)+2), cause (sin(x)+1)^2 - 1(^2) = (sin(x)+1-1)*(sin(x)+1+1)

[manuelodabashian]

Can you do a video on bernoulli numbers? But simple!

[rickpala_]

where are you from? i love your videos :)

[AlgyCuber]

first of all, what do you solve for? x? y? dy/dx?

[dilekyildirim]

Hello I'm turkish. My english is bad but there is a my question for dif equ.this question, what time you do variation of parameter dif equ video. Thanks for video and sorry for my english

[joluju]

Strange scenario : a physicist asks a mathematician for a general solution, the mathematician gives it and leaves, then the physicist makes an experiment and gets an initial condition. From it he finds 2 possible values for C. Who made a mistake ?

[sheepman]

That tutor guy can go to *%&$ for all I care. I got black pen red pen!