integrating factor

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ln 2 = 1 - 1/2 +1/3 - 1/4 ...

use 1) sum of (x ^ (n +1))/ (n +1 ) converges uniformly on [-1, 0] 2) sum of x ^ n converges uniformly on (-1, 0] 3) sum of x ^ n = 1/(1-x) to show that ln 2 = 1 - 1/2 + 1/3 - 1/4 ...

Minimizing question

The illumination at a point is inversely proportional to the square of the distance of the point from the light source and directly proportional to the intensity of the light source. If two light sources are 30 feet apart and their intensities are 70 and 50 respectively, at what point will the sum of their illumination be minimu ...continues

DIFFERENTIAL EQUATIONS

Solve for y in terms of t. Initial conditions are given. We are to use y=Ce^pt form. I tried to do this one: dy/dt=6t^2 y(0)=5 by saying that dy=6t^2dt or that y'=12t and put it into y=Ce^12t^2 by saying that p=12t, then I put in initial condition of 5=Ce^12(0)^2=C to see that C=5, thus y=5e^12t^2. We are to use this meth ...continues

Differential Equations

Solve for y in terms of t. Initial conditions given. a.)dy/dt=20cos5t+8 y(0)=10 b.)dy/dt+50y=0 y(0)=20

Differential Equations

Solve y in terms of t - initial conditions given. a.) (d^2)y/dt^2+7dy/dt+12y=0 y(0)=20 y'(0)=0 b.) (d^2)y/dt^2+4dy/dt+3y=30 y(0)=20 y'(0)=12

Differential Equations

Solve y in terms of t with initial conditions given. a.) (d^2)y/dt^2+3dy/dt+2y=24e^-4t y(0)=10 y'(0)=5 b.) (d^2)y/dt^2+6dy/dt+9y=0 y(0)=10 y'(0)=0

Graphing usng First and second derivatives

Let x^2/(4x+3) a. The graph has vertical asymptotes along the lines x=a for a= b. The horizontal asymptote is y= c. As x approaches a from the left, y approaches c= d. As x approaches a from the right, y approaches d= e. The graph has a local maximum at x= f. The graph has a ...continues

Diff. EQ

Solve for variable y in terms of t W/ given initial condition: dy/dt + 4y = 40sin3t y(0)=6

Diff. EQ

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