it doesn't make sense to me, because how can you add numbers to a smaller number, and get the smaller number for your answer?

What the hell is an integers. https://www.nopistons.com/forums/pub...1047683561.gif

an integer is a whole number, I think. Like, no decimals or fractions, or imaginary numbers.

negative

yeah, I've thought of that, but it still doesn't seem like it could work https://www.nopistons.com/forums/pub...1047683561.gif



I think they may have screwed up writing the problem or something...unless someone has an answer that I can't seem to find

easy



-3 -1 1



-3 + -1 = -4



-4 + 1 = -3

AN interger is non negative because it lies in the set of real numbers. If you go into negative numbers with intergers you are now entering the set of imaginary numbers. Those problems are brain teasers and the way to solve it by trial and error using odd and even numbers. I just failed my math test for Differential Equations.



SOlve this problem:



x^2 means x squared okay..

y' means first derivative.. '' means second and so on

e= eulers equation = 2.7

ln= natural log

cos t^2 + sin t^2 =1 ( identity)

cot = cotangent

sec= secant



IIGht



Find a Particular Solution for



y'' + 4y= 3 csc t ( csc is the cosecant so its 1/sin (t) )



SO the general solution is yc(t) = C1 cos 2t + C2 sin2t



The general solution is off this form because the discriminat b2-4ac ( part of the quadratic equation of the form -b +- (b2-4ac)^(1/2) / 2(a) ). is less than 0. so then the general equation is y= e ^ lambda (t) ( c1 cos u(t) + c2 sin u(t) ) and the discrimant therfore gives us lambda = 0 so e raised to the zero is 1 and then we get C1 cos 2t + C2 sin2t.





Find the coefficiants c1 and c2 byusing variation of parameters where the first condition that needs to be satisfied is



1) v1'y1 + v2'y2= 0

2) v1'y1' + v2'y2' = csc (t) by using trigonometric substituition. Engineers of NP.com gimme this answer!







The answer is y= 3 sint + 3/2 ln | csc t - cot t|sin2t + c1 cos2t + c2 sin2t

rigggggghhhhhhhhhhhhhhhttt





but an integer is still negative, unless they say a real number...then its not negative

math = teh suxx0rz