What is the next level of math after calculus or differential equations?
After the calculus series of courses you should then study linear algebra and introduction to ordinary differential equations. It turns out that linear algebra and ODEs are very connected to each other. Solutions to ODEs form vector spaces. Solving a linear system of ODEs is connected to the eigenvalues and eigenvectors of the corresponding matrix.Then once done with linear algebra and ODEs there are plenty of paths that can be taken.Real analysisHere you revisit the ideas underlying calculus, and now you go through it leaving no stone unturned. Definitions and theorems dealing with limits of sequences, limits of functions, types of continuity, derivatives, integrals, and so on are covered. It took a full semester before I reached the fundamental theorem of calculus.A second semester then typically includes functions of more than one variable, and the generalizations of things from the first semester, metric spaces, normed vector spaces, basic point set topology and so on.Complex analysisThis is the study of functions of a complex variable. Ideas such as differentiating and integrating complex functions. The theory of holomorphic and meromorphic functions. Fractional linear (Moebius) transformations and so on. A lot of techniques of integration of real valued functions make more sense here.Numerical analysis and methodsThe study of algorithms, and their use in solving mathematical problem. Numerical methods can be used to find roots of functions, solve linear systems of equations, interpolate / extrapolate functions, and solve differential equations. This field is closely connected with computability theory.Abstract algebraAlso known as modern algebra, or just plain algebra. Here you study collections of objects more general than the numbers you are used to. Monoids, groups, rings, fields and so on are covered along with their theorems and proofs. This field of mathematics is closely connected with number theory, algebraic geometry, algebraic topology, and cryptography.Probability and StatisticsRandom variables, expected values, variance, correlation, data analysis and so on. A rigorous approach usually comes later with measure theory.In my studies I then also took courses in partial differential equations, optimization theory, functional analysis, measure theory, topology, manifold theory, numerical linear algebra, numerical PDEs, and others.
Calculus 2 problem. Linear Equations? Solve the differential equation.?
4x^3y+x^4y' = sin^3x by inspection the left-hand side is simply the derivative of (x^4y) with respect to x d[x^4y ]/dx = 4x^3*y + x^4* dy/dx = 4x^3y + x^4y' 4x^3y+x^4y' = sin^3x d[x^4y ]/dx = sin^3x [x^4y ] = ∫ sin^3x dx for ∫ sin^3x dx use the reduction formula(in stead of spending hours and hours) ∫ sin^3x dx = -2/3 cos(x) - 1/3 cos(x) sin^2(x) = -2/3 cos(x) - 1/3 cos(x) [ 1 - cos^2x] + C = - cos(x) + 1/3 cos^3x + C = cosx [ 1/3 cos^2(x) - 1 ] + C = cosx [ cos^2(x) - 3 ]/3 + C = 1/3 cosx [ cos^2(x) - 3 ] + C [x^4y ] = ∫ sin^3x dx [x^4y ] = 1/3 cosx [ cos^2(x) - 3 ] + C y = 1/(3x^4) cosx [ cos^2(x) - 3 ] + C/x^4
What should I take after Calculus 3: differential equations or linear algebra?
Differential equations make more sense if you have the notion of vector spaces and linear combinations down.When I was a mathematics major as an undergrad I took linear algebra the same semester as Calculus 3, and then afterwards took introduction to differential equations.As an example, suppose you have the differential equation[math]y'' + 5y' + 6y = 0[/math]We can solve this by knowing that exponentials are the eigenfunction of the derivative, and rewriting as a characteristic algebra problem[math](\frac{d^2}{dx^2} + 5\frac{d}{dx} + 6)y = 0[/math][math](\lambda^2 + 5\lambda + 6)e^{\lambda x} = 0 \Rightarrow[/math][math](\lambda + 2)(\lambda + 3) = 0 \Rightarrow[/math][math]\lambda = -2,-3 \Rightarrow[/math][math]y = C_1 e^{-2x} + C_2 e^{-3x}[/math]See those arbitrary constants [math]C_1, C_2[/math]? Any linear combination of those exponential functions is also a solution. You need to add extra conditions such as initial or boundary conditions to get something unique.Later on when you are solving systems of differential equations you will be calculating the eigenvectors and eigenvalues of matrices. The whole concept of eigenvectors, and eigenfunctions is also a linear algebra concept.Without the linear algebra background a lot of the tools that you learn in ODEs don’t make much sense, and you would have to revert to memorization.
Would it be wise to take differential equations and calculus 2 in the same semester?
No, not wise at all! I can’t recommend you taking first calculus II enough. Most laws of nature are written in terms of differential equations so the number of applications in physics and engineering is quite tremendous. It would be such a shame if you memorized formulas just so you could pass with a D and then saw yourself struggle with any material involving ode’s in the future. If you’re really willing to do it, go over to Calculus II, learn every part of the sections titled integration techniques and series and sequences ( the rest of the material is prerequisite knowledge for a course on vector calculus), and practice enough problems, otherwise, you’ll be heading into a class missing at least half a semester’s worth of math knowledge. You’re not only expected to know this stuff, you’re also expected to be pretty automatic with it.
Differential equation problem: how do you solve: y''=3yy' ?
Note we can rewrite the equation y'' = d/dx (3/2 y^2) Integrating both sides dy/dx = 3/2 y^2 + C1 dy/(3/2 y^2 + C1) = dx Integrating both sides we get: Arctan(sqrt(3/2) y/sqrt(C1))/(sqrt(3/2) sqrt(C1)) = x + C2 sqrt(3/2) y/sqrt(C1) = tan(sqrt(3/2) sqrt(C1) (x + C2)) y = sqrt(C1)/sqrt(3/2) *tan(sqrt(3/2) sqrt(C1) (x + C2)) Cleaning up a bit by redefining our constants: A1 = sqrt(C1)*sqrt(3/2) A2 = A1 * C2 We have as a solution: y = (2/3) A1 tan(A1 x + A2) UPDATE: In case you don't believe me, check (don't screw up the chain rule): y = (2/3) A1 tan(A1*x+A2) y' = 2/3 A1^2 sec^2(A1*x+A2) y'' = 4/3 A1^2 sec(A1*x+A2)*sec(A1*x+A2) *tan(A1*x+A2)*A1 =3*2/3 A1^2*sec^2(A1*x+A2)* 2/3*A1*tan(A1*x+A2) = 3*y'y.
Differential Equation; Calculus 2?
A country's infrastructure is its transportation and communication systems, power plants, and other public institutions. The Solow Model asserts that the value of a national infrastructure K increases due to investments and decreases due to capital depreciation. The rate of increase due to investment is proportional to national income, Y. The rate of decrease due to depreciation is proportional to the value of existing infrastructure. Write a differential equation for K. Please help, i have no idea where to go from here?
-bv^2 force problem! (Solving differential equations!)?
From Newton's second law, we know that: F = m*a = m*v' = m*x'' You are told that F = -b*v^2, so let's first solve for v(t): m*v' = -b*v^2 dv/dt = (-b/m)*v^2 This is a separable equation: dv/v^2 = (-b/m) dt -1/v = -b*t/m + c where c is the constant of integration. v(t) = 1/(b*t/m - c) We are told that v(0) = Vo, so: Vo = 1/(-c) c = -1/Vo v(t) = 1/(b*t/m + 1/Vo) = m*Vo/(Vo*b*t + m) Now integrate this to solve for x(t), after recognizing that v(t) = dx(t)/dt and that the resulting equation is separable: dx/dt = m*Vo/(Vo*b*t + m) dx = m*Vo/(Vo*b*t + m) dt x(t) = (m/b)*ln(Vo*b*t + m) + d where d is a constant of integration. We are told that x(0) = 0, so" x(0) = 0 = (m/b)*ln(m) + d d = -(m/b)*ln(m) So the solution to this initial value problem is: x(t) = (m/b)*ln(Vo*b*t + m) - (m/b)*ln(m) x(t) = (m/b)*ln((Vo*b*t + m)/m) x(t) = (m/b)*ln(1 + Vo*b*t/m)
Do I need Calculus I to III knowledge for differential equations?
The short answer that differential equations (ordinary and partial), not to mention integral equations, are calculus at their root, and require many of the base skills skills taught in Calculus I, Calculus II, and Calculus III.For instance, in integral calculus when you are presented with the following integral[math]\displaystyle F(x)=\int_{\xi=0}^x \xi^2 \sin{\xi} d\xi[/math]can be construed as the following differential equation:[math]F’(x) = x^2 \sin{x}[/math][math]F(0) = 0[/math]A differential equation would allow some generalization of the above, for instance to[math]F’(x)-x F(x) = x^2 \sin{x}[/math][math]F(0) = 0[/math]It should not be surprising the tools of calculus come to bear to solve such problems.When you get to partial differential equations, you not only need Calculus III but linear algebra.
Solve the differential equations?
1) (t^2)y' + 2ty = sin^2(t) y' + 2y/t = sin^2(t)/t^2 P(x) = 2/t, Q(x) = sin^2(t)/t^2 IF = e^[ 2∫dt/t] = e^[2ln(t)] = e^[ln(t)^2] = t^2 yt^2 = ∫ t^2 * sin^2(t)/t^2 dt yt^2 = ∫ sin^2(t)dt yt^2 = 1/2 ∫dt - ∫1/2cos(2t)dt yt^2 = 1/2t - 1/4sin(2t) + C y = 1/(2t) - [2sin(t)cos(t)]/(4t^2) + C/t^2 y = 1/(2t) - [sin(t)cos(t)]/(2t^2) + C/t^2 answer// 2.) y'- 2ty = (3t^2)(e^t^2) P(x) = - 2t, Q(x) = 3t^2e^(t^2) IF = e^[ -2∫tdt] = e^(- t^2) ye^(-t^2) = ∫ e^(-t^2) (3t^2)(e^t^2)dt ye^(-t^2) = 3∫t^2dt ye^(-t^2) = t^3 + C y = t^3e^(t^2) + Ce^(t^2) answer//