Differential Equations with Forcing: Method of Undetermined Coefficients

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This video introduces external forcing to linear differential equations, and we show how to solve these equations with the method of undetermined coefficients. The idea is simple: 1) solve the unforced, or "homogeneous" system; 2) find a particular solution that equals the forcing when plugged into the ODE (generally, we guess the form of the solution based on the form of the forcing); 3) combine the homogeneous and particular solutions and use this full solution to identify the unknown coefficients with initial conditions.
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This video was produced at the University of Washington
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0:00 Overview and problem setup
7:02 Step 1: Solve homogeneous differential equation
8:45 Step 2: Solve for the particular solution
13:50 Step 3: Solve for coefficients with initial conditions

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7 авг 2024

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Комментарии : 11

@YassFuentes Год назад

Marvelous! It's a pleasure to watch your videos.

@alicanakca8116 Год назад

That was flawless, sir. Thanks a lot!

@MrHaggyy Год назад

Great explenation. Much more compact than what i heard at school or university. They usually start with a special case of single sine or cosine expression. Build up to A sin(wt)+ B cos(wt) with a last minute turn to e functions over Eulers formula. Where they usually lose everybody in the hurry. This general picture/strategy tailored down to the special case works way better for me now. ^^ but i hold an Eng. degree now so this stuff is not new for me anymore.

@APaleDot Год назад

Since sin and cos are basically just special cases of the exponential function, it's definitely simpler to start there. Exponentials are easier to conceptualize when using derivatives anyway.

@sam08090 Год назад

Thank you 💗

@mohcinechraibi2605 Год назад

Great lecture! I'm particularly interested to know more about your technical setup. Is this a transparent whiteboard? But then, the writing is not reflected like in a mirror! ^^

@AlistairLynn Год назад

Write on glass and then flip the recorded video horizontally to get the text back in the right direction

@mohcinechraibi2605 Год назад

@@AlistairLynn That makes sense if he is only writing on glass. However, in some other lectures, he also projects python code on the glass.

@hoseinzahedifar1562 Год назад

Very thanks ❤❤❤...As I learned from the problem in 19:01, the values of c1, c2, A, and B depend on the value of \omega. is that right? It is my solution for A and B, when I put x(t) = Acos(wt)+Bsin(wt) in the ODE: [A;B] = [2-w^2 3w; -3w 2-w^2]^-1 * [1;0]

@APaleDot Год назад

So because differentiation is a linear operator, the homogenous solution is like the nullspace of that operator. Just like with linear transformations, the full set of solutions is given by the particular solutions plus the nullspace, because the transformation collapses the nullspace to nothing.