ODE | Initial value problems for second order equations

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Examples and explanations for a course in ordinary differential equations.

We define and solve an initial value problem for a second order linear differential equation, using solutions found earlier. We also discuss the idea of being able to solve *any* initial value problem.

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Hey great set of videos, go to know about a lot of the basics. However is this the end of your course? sadly the end this video is essentially where my course starts.

arianabedi

Thank for this. I was STARVING for math.

SweatyMathMajor

at 3:41, solution to ivp should be y(x) = a cos(x) + b sin(x), right?

rlhugh

When you plug in your initial value for the derivitive function, can you plug in your c1 value you found previously? In this case it did not matter because of the zero cancelling it out, but, of course that's not always going to be the case. IT's still the same c1 correct?

erinmusarra

How come for the solution you put y(x) =cos x +sin x instead of y(0) equals that. Since that is the solution when the initial value is 0?

danielprieto

Can this exact method work when the right side doesn't equal 0?

DoughyBoy

how do you solve for it when one constant doesn't cancel out easily as in this example...

jwpla

Excellent presentation and construction of the video.

heruilin

this was awesome, can we also have pde's

sigmatau

Thanks for the great video Mr. Thompson! It was very helpful. And for the day off from class haha

TylerRealMcNeal

Haha we as a class viewed it as a day off! And sounds good, looking forward to the chapter 7 stuff.

TylerRealMcNeal

Anyone know any videos where one half of the equation isnt conveniently canceled out?
I have no idea how to do this when i have to solve c1 and c2 at the same time.

Literally every video I've found does this :(

Ilcatz

Hmm, I got both c1 and c2 to be [ sqrt(2)/2 * (a-b) ].

Karinia

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