Help solving these differential equations please?
1. Use the method of undetermined coefficients to solve the following differential equation: y '' + 4y = 4x. Using the annihilator approach, I changed the equation to m^2 + 4 = 0, then getting m = 2i, -2i so my equation then became y = c1 cos 2x + c2 sin2x. My annihilator for the other side was D^2. After all my steps, my final answer was y = 4+ x + c1 cos2x + c2 sin2x. Can someone explain what the final answer is and what the correct one is please? 2. Find the solution of y '' - 6y' + 9y = 288 e^9t with y(0) = 3 and y'(0) = 5. Thanks!
How to solve these Differential equations?
The equation of a curve is Y= 2cosX+sin2X Find the x-coordinates of the stationary points on the curve for which 0
How do you solve this differential equation: y' = 1000 + 0.03y?
We need to separate the variables x and y. This becomes clearer if we write it as dy/dx = 1000 + 0.03y. We basically want all x and dx terms on the right, and all y and dys on the left. Now multiply each side by dx: dy = (1000 + 0.03y)dx And divide by the bracketed term: dy/(1000 + 0.03y) = dx. Now we can integrate both sides. int(dx) = x, so the RHS is easy. Now the left. int(dy/(1000 + 0.03y)) = int(1/(1000 + 0.03y))dy. We know from tables that int(1/(ax + b))dx = (1/a)*ln | ax + b |. So int(1/(1000 + 0.03y))dy = (1/0.03)*ln | 1000 + 0.03y | = (100/3)*ln | 1000 + 0.03y |. So we now have (100/3)*ln | 1000 + 0.03y | = x + C. Solve for y: ln | 1000 + 0.03y | = (3/100)*x + A where A = (3/100)*C 1000 + 0.03y = Be^[(3/100)*x] where B = e^A 0.03y = Be^[(3/100)*x] - 1000 y = (1/0.03)*Be^[(3/100)*x]- 1000/0.03 = (100/3)*Be^[(3/100)x] - 100000/3. So if we had an initial condition for y, we could find B and hence our particular solution. e.g. if we had y = y0 when x = 0, then y0 = (100/3)*Be^[(3/100)*0] - 100000/3. = (100/3)*B - 100000/3 So B = (3/100)*(y0 - 100000/3) = ... etc.
Please help me solve this differential equation such that y = pi/3 when x=1?
it is seperable dy/dx = 1/x cot y so tany dy = 1/x dx slip in an integral sign ∫tany dy = ∫1/x dx ln|secy| = lnx + C then y=pi/3 when x=1 gives C = ln2 so lnsecy = lnx + ln2 = ln(2x) cosy = 1/(2x) y = arccos[1/(2x)] ,.,.,.,.
Could u help me solve these 2 differential equation problems?
1.Determine complementary solutions using undetermined coefficients a.d^4/dx^4 + 4(d^2y/dx^2) = 3x+sin2x 2.Determine the laplace transform of the following periodic functions. a.f(t) = [sint 0 ≤ t < p ] [ 0 p ≤ t < 2p ] f(t) = f(t+2p)
How to solve these differential equations please?
1) e^(-4z) dz = 2t dt (-1/4) ∫(-4e^(-4z)) dz = ∫2t dt (-1/4) e^(-4z) = t^2 + C t = 0 --> z = 0 C = - 1/4 (-1/4) e^(-4z) = t^2 - 1/4 multiply both sides by (-4) e^(-4z) = 1 - 4t^2 -4z = ln |1 - 4t^2| z = (-1/4) ln |1 - 4t^2| 2) dy/dx = x^5 y dy/y = x^5 dx ln y = (x^6)/6 + C y = ke^((x^6)/6) y(1) = 2 ke^(1/6) = 2 k = 2e^(-1/6) y = 2e^(-1/6) e^((x^6)/6) y = 2e^((x^6-1)/6)
How can I self study Differential Equations?
Gilbert Strang has done an amazing job making his instruction on several math topics including differential equations widely available for free[1]. He is an incredibly talented teacher. Back when I tutored Diff. Eq. I swore by Paul's online notes, which is an amazing reference for brushing up on topics quickly[2]. Also, check out the sage math website[3], it's a free desktop or online computing platform for solving and plotting problems that runs on a few different scripted language implementations. The key to getting your mind around diff. eqs. is to plot, plot, and plot again (at least for me). Good luck and enjoy!Footnotes[1] http://math.mit.edu/~gs/dela/[2] Differential Equations[3] SageMath Mathematical Software System - Sage
What is the Solution of this differential equation?
The quotient rule d/dx(x/y) = [y -x(dy/dx)]/y^2Divide the eqn. by y^2[y-x(dy/dx ) ]/y^2 = cos x - sin x .cos xd(x/y)/dx = cos x - (1/2)sin(2x)d(x/y) = cosx dx - (1/2) sin 2x dx(x/y) = sin x +(1/4) cos 2x +cy = 4x/(4 sin x +cos 2x +k)