if m1 and m2 are different numbers. Which one should pick for c1 exponent and which one for c2 exponent. I mean take the bigger number for c1 e and smaller number for c2 e? like example 2?
It's still the same fundamental idea. You just have a characteristic cubic (or higher degree polynomial), instead of a characteristic quadratic to solve. The idea is still to solve the equation assuming it's homogeneous, by assuming a solution of e^(r*t), and differentiating to match the original diffEQ. Then we solve for r, and construct an arbitrary linear combination of e^(r*t) based on all possible values of r. Similar procedures also apply for solving non-homogeneous diffEQs of higher order, by finding the particular solution and combining it with the homogeneous part.
You say "linear, because the coefficients are functions of x". But that is not the reason for the linearity of the DE. Even if the coefficients were not functions of x this equation would be a LDE. It is linear because the equation is a linear polynomial in the unknown function and its derivatives, i.e., the unknown function and all its derivative terms only appear to the maximum power of 1 (see also here en.wikipedia.org/wiki/Linear_differential_equation).
Very slow, calm and clear. ❤
That's great
@@SkanCityAcademy_SirJohn Good morning sir. Please sir am stock to solve.
y"-2y'+y = ((3x^2)+1)e^x
Am stuck on the particular integral.
Thanks
Am letting Yp = (ax^2+bx+c)e^x.
And it is giving.
a=1/2, b(disappeared), c(disappeared).
Please help me out❤
Okay, so once ax2.... Fails, you then try ax3... In that other until it matches up
Okay sir thanks very much.
So there is no way to know what level to go..?
ax^n at once?
I enjoy your videos a lot 🇿🇲🇿🇲
Aww that's nice
thank you very much ❤
You are most welcome
Very nice, thank you sir
you are most welcome
if m1 and m2 are different numbers. Which one should pick for c1 exponent and which one for c2 exponent. I mean take the bigger number for c1 e and smaller number for c2 e? like example 2?
You can use any of them
Good teacher
Thanks so much
Thank you Sir.
You are most welcome
You are interesting me and also I need to know how to solve first order differential equations of higher degree
It's still the same fundamental idea. You just have a characteristic cubic (or higher degree polynomial), instead of a characteristic quadratic to solve. The idea is still to solve the equation assuming it's homogeneous, by assuming a solution of e^(r*t), and differentiating to match the original diffEQ. Then we solve for r, and construct an arbitrary linear combination of e^(r*t) based on all possible values of r. Similar procedures also apply for solving non-homogeneous diffEQs of higher order, by finding the particular solution and combining it with the homogeneous part.
Nice contribution.
pls I'm looking for higher order differential equations do you have it on your playlist?
It's up to 2nd order DE
İf bita = 3/4 i for example, are we going to use bita=3/4
Yes
Atleast i've gain something🥰
Thank you so much
You say "linear, because the coefficients are functions of x". But that is not the reason for the linearity of the DE. Even if the coefficients were not functions of x this equation would be a LDE. It is linear because the equation is a linear polynomial in the unknown function and its derivatives, i.e., the unknown function and all its derivative terms only appear to the maximum power of 1 (see also here en.wikipedia.org/wiki/Linear_differential_equation).
Thanks for your contribution, that is a great enlightenment. Thanks so much. Much love❤️
It's waaaa
Thanks so much bro