Correction: You meant to write “homogeneous” (note the extra “e”) in the title. “Homogenous” is an old-fashioned term from biology meaning similar in structure but not necessarily function (like the human arm and the wing of a bat). [you said it correctly in the video, but that’s not the word in the title]
It is a linear differential equation of the form dy/dx+P(x)y=Q(x) Here P(x)=1,Q(x)=x³ Thus integrating factor=e^integral(1)dx=e^x General solution:y(IF)=integral (Q(x)(IF)dx y e^x=integral e^x(x³)dx y=x³-3x²+6x-6+c
@@holyshit922 I made mistake The general solution of Linear differential equation of first order and first degree of that form is y (IF)=integral(Q(x) (IF) dx)+c Thus y=x³-3x²+6x-6+ce^-x
We have three main types of first order ordinary equations which can be found in textbooks -separable ode -linear ode -exact ode Then we have three methods of reducing one type of equation to another - substitution (f. e. homogeneous to separable , Bernoulli to linear) - integrating factor (Method of choice for linear ode but also allows to reduce equation to exact. Generally difficult to find but there are some special cases in which it is relatively easy to find it) -introducing parameter (Lagrange equation and its special case Clairaut , yes Alexis is a male name) Your equation is first order linear and non-homogeneous
Correction: You meant to write “homogeneous” (note the extra “e”) in the title. “Homogenous” is an old-fashioned term from biology meaning similar in structure but not necessarily function (like the human arm and the wing of a bat).
[you said it correctly in the video, but that’s not the word in the title]
Thank you! Fixed
It is a linear differential equation of the form
dy/dx+P(x)y=Q(x)
Here P(x)=1,Q(x)=x³
Thus integrating factor=e^integral(1)dx=e^x
General solution:y(IF)=integral (Q(x)(IF)dx
y e^x=integral e^x(x³)dx
y=x³-3x²+6x-6+c
Yes , for me linear in the word missing in the title
Why -6 + c ?
@@holyshit922 Added!
@ He missed exp(-x)
@@holyshit922 I made mistake
The general solution of Linear differential equation of first order and first degree of that form is y (IF)=integral(Q(x) (IF) dx)+c
Thus y=x³-3x²+6x-6+ce^-x
Nice problem that i solved in my head😊💯💥
Me too!
We have three main types of first order ordinary equations which can be found in textbooks
-separable ode
-linear ode
-exact ode
Then we have three methods of reducing one type of equation to another
- substitution (f. e. homogeneous to separable , Bernoulli to linear)
- integrating factor
(Method of choice for linear ode but also allows to reduce equation to exact.
Generally difficult to find but there are some special cases in which it is relatively easy to find it)
-introducing parameter (Lagrange equation and its special case Clairaut , yes Alexis is a male name)
Your equation is first order linear and non-homogeneous
(D+1)yp=x^3,yp=(D+1)^-1(x^3)=(1-D+D^2-D^3+...)(x^3)=x^3-3x^2+6x-6. yh= ke^-x, y = ke^-x+x^3-3x^2+6x-6.
y = x^3 + ax^2 + bx + c
y’ = 3x^2 + 2ax + b
y + y’ = x^3 + (a + 3)x^2 + (b + 2a)x + (b + c) = x^3
y = x^3 - 3x^2 + 6x - 6
This method: similar to solving second order linear ODE
y=yo+yp=ce^(-x)+x^3-3x^2+6x-6
I like your pace. Not boring at all. Another podcaster works very slow with lots of in-between steps making me very sleepy watching his presentations.
Glad to hear that! Thanks for the feedback
(ye^x)' = x^3e^x
ye^x = ∫x^3e^x dx = e^x(x^3 - 3x^2 + 6x - 6) + C
y = x^3 - 3x^2 + 6x - 6 + Ce^(-x)
(e^x*y)'=e^x*x³
e^x*y=∫(e^x*x³)dx=(x³-3x²+6x-6)e^x+C
∴y=C*e^(-x)+x³-3x²+6x-6
Nice!
Still quipping on the 2b? Seriously mate, just stop.
It would be more elegant to solve for
y(x) = P₁(x)•e^P₂(x)
how?