A Quick and Easy Polynomial Differential Equation

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#Calculus #DifferentialEquations

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[richardryan]

The given equation is a first order linear differential equation. There is a formula for solving 1st order linear differential equations that is presented in an Intro to D.E.s course which, oftentimes, is introduced in a Calculus II course. If the restriction that P(x) must be a polynomial is not included, then the general solution is P(x) = x^2 + 2x + 2 + C*e^x, where C is a constant.

[MINEXKILLER]

IDK why but when comparing coefficients and making a system of equations and solving for them is so satisfying

[shamanjitsingh]

I solved using the integrating factor method as per Bernoulli's equation. I got answer as P(x)=x^2+2x+2+ce^x.
For this function to be polynomial c = 0 and we get our answer.

[diogenissiganos]

This video is easy to understand because it doesn't involve too much calculus!

[satyapalsingh]

Your method of solving the differential equation is unique .

[chimmychonga]

I found a faster way of doing it

Define p(x)=x^2+p'(x)
Apply the derivative to both sides and we get
p'(x)=2x+p"(x)
Keep pulling this trick and substitute the derivative values in
Since p(x) is at most a quadratic, then continuing to add will just give zeros
We get that same answer

[sngmn]

for y = y(x) and y - y' = x^2
y = Ce^x + x^2 + 2x + 2
if C = 0 then P(x) = x^2 + 2x + 2.

[johnchristian]

Just did differential equations in my course, nice!

[math]

love the polynomial solution bro, great job

[aparslan]

This was easier then other videos imo.

[MathElite]

Wow a differential equation on this channel, wasn't expecting that
You want to solve P(x)-P'(x)=0 for one part
then you use substitution to get the other part, then you add them

[amitshoval]

Gret video! Make more about differential equations please!

[raystinger]

(before watching solution) So let's consider P(x) is a polynomial of degree N, therefore P'(x) is a polynomial of degree N-1.
P(x) - P'(x) = x²
=> P(x) = P'(x) + x²
Now looking at this equation, we have a polynomial of degree N on the left hand side, therefore the polynomial on the right side must have the same degree. However, if N-1≥2, the polynomial in the right hand side will have degree N-1, and since N≠N-1, therefore N-1<2. Also, if N≤1, then the right hand side will have degree 2 while the left hand side will have a lower degree. Therefore N must be equal to 2.
So let's consider P(x) = ax² + bx + c
=> P'(x) = 2ax + b
=> P(x) = P'(x) + x² = 2ax + b + x² = ax² + bx + c
=> a = 1, b = 2a and c = b
=> a = 1, b = 2 and c = 2
=> P(x) = x² + 2x + 2

[titassamanta]

Very easy. Learnt differentiation last month (though I discovered some formulas for the areas under the graph of polynomials and also found out the volume formulas of a sphere and cone last year on my own using the method of exhaustion. Later my father told me that if was basically integration and insisted that I start learning calculus.)

[BCS-IshtiyakAhmadKhan]

Just multiply both sides by (-e^-x) then integrate

[mvashishtha]

The explanation given that it is quadratic equation is vague this however can be shown
P(x)- P(-x)= an * x^n + (an - n * an-1)* x^n-1+ ....
an = 0, an - n * an-1 = 0...
an-1 = an / n = 0
...
a3 = a4 / 3 = 0
a2 = 1
a1 = 2 * a2=2
a0 = a1=2

[manojsurya]

Dude it was a super creative approach! I know u always wouldn't go with the standard way (integrating factor method)😃

[gastonsolaril.]

"f(x) = a e^x + x^2 + 2 x + 2" but with "a = 0"... XD
I would really like for you to include more differential equations, but the "rare" ones!

[Vtalt]

Using a differential operator (D*p=p’): p-p’=p(1-D)=x^2 =>
Don’t know why it works though)) my teacher just told me “trust me on that one, it always does work with polynomials”

[aashsyed]

You are so awesome 😎😎😎 and you are a pro! A genius! Your channel is growing fast!!!