given in series circuit L=1H, R=110ohm, C=0.001F, v=90V. Find the resultant current when the switch is originally open and it is closed for 1s. then it is open again when t=1. use laplace transform...
given in series circuit L=1H, R=110ohm, C=0.001F, v=90V. Find the resultant current when the switch is originally open and it is closed for 1s. then it is open again when t=1. use laplace transform...
find the particular solution of the differential equation f"(x)=6(x-1) whose graph passes through the point (2,1) and is tangent to the line 3x-y-=0 at that point
The process I was taught in class was as follows: 1. Transform the second order non homogeneous DE into a system of first order DE. 2. Solve the homogeneous part. 3. Assume the solution...
Attached is the problem from my text book. I'm confused about what I'm actually solving and how to use what I know. I know that Current for the inductor is given by the equation di/dt = v/L and the...
Suppose that an initial population of 10,000 bacteria grows exponentially at a rate of 2% per hour and that y=y(t) is the number of bacteria present t hours later. a) Find an initial-value...
how do you find the second derivative of y=e-x^2
after the coffee is heated to 200 degrees , it is set in a room at 50 degrees.It cools at a rate that is proportional to the difference between its current temperature and that of the surrounding...
A differential equation is an equation involving a function and one or more of its derivatives. Determine whether the function y=πsinθ+2πcosθ is a solution to the differential equation d^2y/dθ^2...
Answer: c1e^-t + c2t^-t + t^2e^-t
consider the equation f(x)=-6-1 and g(x)=4x^2 select the solution for (fg)(x)
Answer: c1e^-t + c2e^-t/2 + t^2 +6t + 14 -3/10sin(t) - 9/10cos(t) Please show the work step by step.
Answer: c1cos(t) + c2sin(t) -1/3tcos(2t) -5/9sin(2t) Please show the work step by step.
Solve the differential equation (2xy+y^3)dx+(x^2+3xy^2-2y)dy=0. I got x^2*y+xy^3+yx^2+xy^3-y^2=C but that's not the answer. And is Differential Equations taught after Linear...
Consider the initial value problem: y'+(2/3)y=1-(1/2)t, y(0)=yο. Find the value of yο for which the solution touches, but does not cross, the t-axis.
Solve the differential equation y^(4)-5y"-36y=0. r^4-5r^2-36 (r^2-9)(r^2+4) r=3, -3, 2i, -2i y=c1*e^3x+c2*e^-3x+c3(cos(2x))+c4(sin(2x)) Is that the right answer...
the position function s of a point is given by two dimensional quantities where t in seconds and the y direction is affected by gravity (10m/s/s). y(t)=4t*sin(θ) - 5t2 ...
Find the general solution of ty'-y=t^2*e^(-t), t>0. I got y=(t^-1)(e^-t)(t^3/3+c) but when I simplify that I didn't get the right answer. Please help me with steps.
Solve the differential equation y"+4y=0. r^2+4=0 r=2i, -2i Now what?
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Solve the differential equation e^x dx+(e^x cot(y)+2y csc(y))dy=0. M=e^x N=e^x cot(y)+2y csc(y) My=0 Nx=e^x cot(y) Obviously isn't exact. What to do next?