both SEPARABLE & LINEAR differential equation

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Part1 of Differential Equation Course: How to solve first order differential equations? The topics/technique include:

separable differential equations,
first order linear differential equations,
integrating factor,
exact differential equations,
special integrating factor,
solve differential equations by substitution,
homogeneous differential equations,
bernoulli differential equation,

(this playlist provides the ideas behind each kind of diff eq, the strategies to solve each kind of diff eq, and plenty of examples/homework examples)

YOU CAN'T SOLVE DIFFERENTIAL EQUATION WITHOUT KNOWING YOUR INTEGRALS!

blackpenredpen
separable differential equations,
first order linear differential equations,
integrating factor,
exact differential equations,
special integrating factor,
solve differential equations by substitution,
homogeneous differential equations,
bernoulli differential equation,

(this playlist provides the ideas behind each kind of diff eq, the strategies to solve each kind of diff eq, and plenty of examples/homework examples)

YOU CAN'T SOLVE DIFFERENTIAL EQUATION WITHOUT KNOWING YOUR INTEGRALS!

blackpenredpen

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The background music cures my depression

shobhitmishra

Thank you for an awesome video on both Separable and Linear Differential Equations.

georgesadler

Elementary School: *_"S h O W Y o U R W O r K!"_*
Middle School: *_"s H o w y O u r w o R k!!"_*
High School: *_"S H O W Y O U R W O R K!!!"_*

bprp: *"the best way you can do these kinds of things is look at this and do the integral - do the U-Substitution - in your head"* 🙂🙃

isaackay

Thanks for the vid! I'll be starting my differential equations class this semester.

scottryder

do you have videos for eliminating arbitrary constant?

shaunhugo

Why bother using a integrable factor? Can't you just use substitution?

y' = x^3*(1-y)

u = 1 - y
y = 1 - u
dy/dx = -du/dx
y' = -u'

- u' = x^3*u
u' = -x^3*u
u'/ u = -x^3
ln u = -x^4/4 + c_1
u = e^(-x^4/4 + c_1)
u = e^(-x^4/4)*e^c_1

y = 1 - u
y = 1 - e^(-x^4/4)*e^c_1

e^c_1 = c
y = 1 - c*e^(-x^4/4)
3 = 1 - c*1
2 = -c
c = -2
y = 1 + 2e^(-x^4/4)

The only weirdness lies in trying to solve for c_1. e^c_1 = -2. I suppose c_1 can be complex, but I don't think integration prohibits imaginary constants.

tsujimasen

both SEPARABLE & LINEAR differential equation

both SEPARABLE & LINEAR differential equation

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