Massachusetts Institute of Technology professor, Gilbert Strang, explains differential equations and linear algebra which are two crucial subjects in science and engineering. This video series develops those subjects both separately and together and supplements Gil Strang's textbook on this subject.

*dy/dt* = *y, dy/dt* = –*y, dy/dt* = *2ty*. The equation *dy/dt* = *y***y* is nonlinear.

*x ^{n}*, sin(

*e ^{st}*, from outside and exponential growth,

*t*) produces an oscillating output with the same frequency ω (and a phase shift).

*q(s)* by its growth factor and integrate those outputs*.*

*f =* cos(ω*t*) is the real part of the solution for *f = e ^{iωt}*. That complex solution has magnitude

*e ^{-at}* multiplies the differential equation, y’=ay+q, to give the derivative of

*–by ^{2}* slows down growth and makes the equation nonlinear, the solution approaches a steady state

*t* and the other in *y*. The simplest is *dy/dt = y*, when *dy/y* equals *dt*. Then ln(*y*) = *t + C*.

*f* = cos(ω*t*), the particular solution is *Y**cos(ω*t*). But if the forcing frequency equals the natural frequency there is resonance.

*e ^{st}*. The exponent

*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.

*y = G* cos(ω*t –* α). The damping ratio provides insight into the null solutions.

*L* (inductance), *R* (resistance), and *1/C* (*C* = capacitance).

*t*, cosines/sines, exponentials), a particular solution has this same form.

*(at ^{2} + bt +c) e^{st}*: substitute into the equation to find

*y _{1}* and

*y(t)*.

*s ^{2}Y* and the algebra problem involves the transfer function

*(t)*, the impulse response is *g(t)*. When the force is *f(t)*, the response is the “convolution” of *f* and *g.*

*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.

*(y, dy/dt)* travels forever around an ellipse.

*y* and *dy/dt*. The matrix becomes a companion matrix.

*Y* to the differential equation *y’ = f(y)*. Near that *Y*, the sign of *df/dy* decides stability or instability.

*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*.

*y’ = Ay* are stable (solutions approach zero) when the trace of *A* is negative and the determinant is positive.

*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}*.

*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.

*A* has a row for every edge, containing -1 and +1 to show the two nodes (two columns of *A*) that are connected by that edge.

**x** remain in the same direction when multiplied by the matrix (*A***x** *=* λ**x**). An *n* x *n* matrix has *n* eigenvalues.

*n* independent eigenvectors. The diagonal matrix Λis the eigenvalue matrix.

*A* = *V*Λ*V ^{–1}* also diagonalizes

*d***y***/dt = A***y** contains solutions **y** *= e ^{λt}*

**y** *= e ^{At}*

*A* and *B* are “similar” if *B* = *M ^{-1}AM* for some matrix

*n* perpendicular eigenvectors and *n* real eigenvalues.

*d ^{2}y/dt^{2} + Sy =* 0 has

^{T}Sv for every vector v. S = A^{T}A is always positive definite if A has independent columns.

*A* into an orthogonal matrix *U* times a diagonal matrix Σ (the singular value) times another orthogonal matrix V^{T}: rotation times stretch times rotation.

*y(0)* and *dy/dt(0)* to boundary conditions on *y(0)* and *y(1)*.

* ^{2}u/*∂

*F(x)* into a combination (infinite) of all basis functions cos(*nx)* and sin(*nx)*.

*F(–x) = F(x)*) and odd functions use only sines. The coefficients *a _{n}* and

*u*(*r*, θ) combines *r ^{n}* cos(

*u*/∂*t* = ∂* ^{2}u*/∂

* ^{2}u*/∂