From the series: Differential Equations and Linear Algebra
Gilbert Strang, Massachusetts Institute of Technology (MIT)
The integral of a varying interest rate provides the exponent in the growing solution (the bank balance).
OK? I want to talk about a slightly different way to solve a linear first-order equation. And if you look at the equation-- I'll do an example. That's the best. Do you notice what's different from our favorite equation? The change is 2t. The interest rate a is increasing with time, changing with time. So we still have a linear equation, still just y. But the coefficient is varying. We have a variable coefficient 2t.
And if we think here of applications to economy, to banks, that would be rampant inflation, the interest rate 2t climbing and climbing forever. But we want to see that this is a class of equations that we can solve. OK. And the new method is called an integrating factor. It's a magic factor that makes the equation simple. So that's another nice way to solve all the problems that we've dealt with so far, plus this new one.
So what is this factor? Well, for this 2t problem, the right factor is e to the minus t squared. And why is that the right factor? This is the factor that I'm going to multiply the equation by and make it simple. And the reason that's the right choice is that the derivative of this-- you remember how to take the derivative by the chain rule?
The derivative will be the same e to the minus t squared, the same I, times the derivative of the exponent. And the derivative of that exponent is minus 2t. Minus t squared becomes minus 2t. So it's that little device that gives us an integrating factor that makes the equation simple.
And now I'm going to look at the equation. What I want to look at is the derivative of I times y. Instead of just dy dt, let me look at the derivative of I times y. So I have a product here. Got to use the product rule. So that will be I-- so I dy dt. OK. dy dt is-- we can take dy dt from the equation, I times 2ty plus q of t. And now I have to add on dI dt y. Good.
So it's the product rule-- I times the derivative of y plus the derivative of I times y. But now look. dI dt we know is minus 2tI. So that dI dt, now I'm using the key fact about I, that that's minus 2tIy. Look, minus 2tIy cancels 2tIy. So I have a nice equation now. The derivative of Iy is Iq. The derivative of Iy is Iq. I can just integrate both sides. And that's the key. That's the key.
If I integrate the left-hand side-- so I'll just move this up-- integrate the derivative-- of course, the integral of the derivative is the function-- at time t, Iy at time t minus Iy at time 0, y of 0. Because notice that I at t equals 0-- can I just mention that-- I at 0 is 1. When t is 0, I of e to the 0 power, which is 1. So that I of 0 is 1. So that's the integral of the derivative.
And on the right-hand side, I have the integral from 0 to t of I times q. So I'll put in-- yeah, e to the minus s squared q of s ds. I've introduced a variable of integration s going from 0 to t. You remember this type of formula? The input is continuous over time, and I'm looking at the resulting output at time t. So all the inputs go in. They're all multiplied by some factor and integrated gives the total result from those inputs.
OK. I'm almost here. I just want to remember I want to divide by I of t so I have a formula for y. OK. So my formula for y. When I divide by I of t-- don't forget what I of t is. Let me put it again here. Let me remind myself. I of t is e to the minus t squared. That was the magic integrating factor.
OK. So I'm going to divide by that, which means I'll multiply by e to the t squared. So that will knock out the I here. I'll put this on the other side of the equation, y of 0, y of 0, and it will be multiplied by the e to the t squared. And this thing will be multiplied by e to the t squared. The integral from 0 to t of e to the t squared minus s squared q of s ds. That's my answer.
Well, let's look at it. I have y of t. This is what comes out of y of 0. You see that the growth factor has changed from our old e to the at-- that was the growth at constant rate, interest rate a-- to e to the t squared. That's our growth from an increasing interest rate.
And over here, I'm seeing the result, the output, from the input q, from all the inputs between 0 and t. Each input is multiplied by now that factor is the growth not from 0 to t. This is the growth from 0 to t. This is the growth from s to t, because the input went in at time s, and it had the shorter time, t minus s, to grow.
So that's the formula for the answer. If you give me any particular q of s, I just do the integral, and I find the solution to the differential equation. So that integrating factor has made things work.
Maybe I should say what the integrating factor would be in general. So let me take a moment to see-- this was an example. This was an example with a of t equal to 2t. What's the general integrating factor? So we always want the integrating factor. Our construction rule is that it should give us-- the derivative should be minus a of t times I itself. That's how we chose the e to the minus t squared. Then a of t was 2t.
Now I want to give the general rule. The general rule for the integrating factor is the solution to that equation. The solution to that equation is giving us the e to the t squared in the example. This was the example.
But now I want a formula just to close off the entire case of varying interest rate. I want to find the solution to that equation. And it is-- so here's the integrating factor. It's e to the minus because of that minus sign. Now I'm wanting a to come down when I take a derivative. So what I'll put up here, the integral of a of t dt, say, from 0 to t.
Now, let me just do again this example just to see. I have e to the minus the integral of 2t, which is e to minus t squared. That's how I get t squared as the right choice for our example. And the general rule is there. That's the integrating factor.
And finally, finally, if a is a constant, which is the most common case-- the only case we've had until this video-- if a is a constant, then the integral of a from 0 to t is just a times t. So number one example, number zero example, would be e to the minus at. That would be the correct integrating factor if we had constant a.
And I'll create some examples, some problems, just to go through the steps in that best case of all with constant integrating factor. But now we can solve it with a varying interest rate. Good. Thank you.
1.1: Overview of Differential Equations Linear equations include dy/dt = y, dy/dt = –y, dy/dt = 2ty . The equation dy/dt = y *y is nonlinear.
1.2: The Calculus You Need The sum rule, product rule, and chain rule produce new derivatives from the derivatives of xn , sin(x ) and ex . The Fundamental Theorem of Calculus says that the integral inverts the derivative.
1.4b: Response to Exponential Input, exp(s*t) With exponential input, est , from outside and exponential growth, eat , from inside, the solution, y(t), is a combination of two exponentials.
1.4c: Response to Oscillating Input, cos(w*t) An oscillating input cos(ωt ) produces an oscillating output with the same frequency ω (and a phase shift).
1.4d: Solution for Any Input, q(t) To solve a linear first order equation, multiply each input q(s) by its growth factor and integrate those outputs.
1.4e: Step Function and Delta Function A unit step function jumps from 0 to 1. Its slope is a delta function: zero everywhere except infinite at the jump.
1.5: Response to Complex Exponential, exp(i*w*t) = cos(w*t)+i*sin(w*t) For linear equations, the solution for f = cos(ωt ) is the real part of the solution for f = eiωt . That complex solution has magnitude G (the gain).
1.6: Integrating Factor for a Constant Rate, a The integrating factor e-at multiplies the differential equation, y’=ay+q, to give the derivative of e-at y: ready for integration.
1.6b: Integrating Factor for a Varying Rate, a(t) The integral of a varying interest rate provides the exponent in the growing solution (the bank balance).
1.7: The Logistic Equation When –by2 slows down growth and makes the equation nonlinear, the solution approaches a steady state y( ∞) = a/b.
1.7c: The Stability and Instability of Steady States Steady state solutions can be stable or unstable – a simple test decides.
2.1: Second Order Equations For the oscillation equation with no damping and no forcing, all solutions share the same natural frequency.
2.1b: Forced Harmonic Motion With forcing f = cos(ωt ), the particular solution is Y *cos(ωt ). But if the forcing frequency equals the natural frequency there is resonance.
2.3: Unforced Damped Motion With constant coefficients in a differential equation, the basic solutions are exponentials est . The exponent s solves a simple equation such as As2 + Bs + C = 0 .
2.3c: Impulse Response and Step Response The impulse response g is the solution when the force is an impulse (a delta function). This also solves a null equation (no force) with a nonzero initial condition.
2.4: Exponential Response - Possible Resonance Resonance occurs when the natural frequency matches the forcing frequency — equal exponents from inside and outside.
2.4b: Second Order Equations With Damping A damped forced equation has a particular solution y = G cos(ωt – α). The damping ratio provides insight into the null solutions.
2.5: Electrical Networks: Voltages and Currents Current flowing around an RLC loop solves a linear equation with coefficients L (inductance), R (resistance), and 1/C (C = capacitance).
2.6: Methods of Undetermined Coefficients With constant coefficients and special forcing terms (powers of t , cosines/sines, exponentials), a particular solution has this same form.
2.6b: An Example of Method of Undetermined Coefficients This method is also successful for forces and solutions such as (at2 + bt +c) est : substitute into the equation to find a, b, c .
2.6c: Variations of Parameters Combine null solutions y1 and y2 with coefficients c1(t) and c2(t) to find a particular solution for any f(t).
2.7: Laplace Transform: First Order Equation Transform each term in the linear differential equation to create an algebra problem. You can then transform the algebra solution back to the ODE solution, y(t) .
2.7b: Laplace Transform: Second Order Equation The second derivative transforms to s2Y and the algebra problem involves the transfer function 1/ (As2 + Bs +C).
3.1: Pictures of the Solutions The direction field for dy/dt = f(t,y) has an arrow with slope f at each point t, y . Arrows with the same slope lie along an isocline.
3.2: Phase Plane Pictures: Source, Sink Saddle Solutions to second order equations can approach infinity or zero. Saddle points contain a positive and also a negative exponent or eigenvalue.
3.2b: Phase Plane Pictures: Spirals and Centers Imaginary exponents with pure oscillation provide a “center” in the phase plane. The point (y, dy/dt) travels forever around an ellipse.
3.2c: Two First Order Equations: Stability A second order equation gives two first order equations for y and dy/dt . The matrix becomes a companion matrix.
3.3: Linearization at Critical Points A critical point is a constant solution Y to the differential equation y’ = f(y) . Near that Y , the sign of df/dy decides stability or instability.
3.3b: Linearization of y'=f(y,z) and z'=g(y,z) With two equations, a critical point has f(Y,Z) = 0 and g(Y,Z) = 0. Near those constant solutions, the two linearized equations use the 2 by 2 matrix of partial derivatives of f and g .
3.3c: Eigenvalues and Stability: 2 by 2 Matrix, A Two equations y’ = Ay are stable (solutions approach zero) when the trace of A is negative and the determinant is positive.
5.1: The Column Space of a Matrix, A An m by n matrix A has n columns each in R m . Capturing all combinations Av of these columns gives the column space – a subspace of R m .
5.4: Independence, Basis, and Dimension Vectors v 1 to v d are a basis for a subspace if their combinations span the whole subspace and are independent: no basis vector is a combination of the others. Dimension d = number of basis vectors.
5.5: The Big Picture of Linear Algebra A matrix produces four subspaces – column space, row space (same dimension), the space of vectors perpendicular to all rows (the nullspace), and the space of vectors perpendicular to all columns.
5.6: Graphs A graph has n nodes connected by m edges (other edges can be missing). This is a useful model for the Internet, the brain, pipeline systems, and much more.
6.1: Eigenvalues and Eigenvectors The eigenvectors x remain in the same direction when multiplied by the matrix (A x = λx ). An n x n matrix has n eigenvalues.
6.2: Diagonalizing a Matrix A matrix can be diagonalized if it has n independent eigenvectors. The diagonal matrix Λis the eigenvalue matrix.
6.3: Solving Linear Systems d y /dt = A y contains solutions y = eλt x where λ and x are an eigenvalue / eigenvector pair for A .
6.4: The Matrix Exponential, exp(A*t) The shortest form of the solution uses the matrix exponential y = eAt y (0) . The matrix eAt has eigenvalues eλt and the eigenvectors of A.
6.4b: Similar Matrices, A and B=M^(-1)*A*M A and B are “similar” if B = M-1AM for some matrix M . B then has the same eigenvalues as A .
6.5: Symmetric Matrices, Real Eigenvalues, Orthogonal Eigenvectors Symmetric matrices have n perpendicular eigenvectors and n real eigenvalues.
7.2: Positive Definite Matrices, S=A'*A A positive definite matrix S has positive eigenvalues, positive pivots, positive determinants, and positive energy vT Sv for every vector v. S = AT A is always positive definite if A has independent columns.
7.2b: Singular Value Decomposition, SVD The SVD factors each matrix A into an orthogonal matrix U times a diagonal matrix Σ (the singular value) times another orthogonal matrix VT : rotation times stretch times rotation.
7.3: Boundary Conditions Replace Initial Conditions A second order equation can change its initial conditions on y(0) and dy/dt(0) to boundary conditions on y(0) and y(1) .
8.1: Fourier Series A Fourier series separates a periodic function F(x) into a combination (infinite) of all basis functions cos(nx) and sin(nx) .
8.1b: Examples of Fourier Series Even functions use only cosines (F(–x) = F(x) ) and odd functions use only sines. The coefficients an and bn come from integrals of F(x) cos(nx ) and F(x) sin(nx ).
8.1c: Fourier Series Solution of Laplace's Equation Inside a circle, the solution u (r , θ) combines rn cos(n θ) and rn sin(n θ). The boundary solution combines all entries in a Fourier series to match the boundary conditions.
8.3: Heat Equation The heat equation ∂u /∂t = ∂2u /∂x2 starts from a temperature distribution u at t = 0 and follows it for t > 0 as it quickly becomes smooth.