Originally Posted by ROTARYROCKET7' date='Jun 23 2003, 08:00 AM

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

Originally Posted by ROTARYROCKET7' date='Jun 23 2003, 08:00 AM

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

Originally Posted by djgiantrobot' date='Jun 23 2003, 12:21 PM

[quote name='ROTARYROCKET7' date='Jun 23 2003, 08:00 AM'] 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

Originally Posted by djgiantrobot' date='Jun 23 2003, 08:21 AM

About a year ago i would have actually tried to help you with that, you'd be shocked how quick you'll loose all that once you are out of school. oh yeah, someone hire me.

Originally Posted by Joe Flo' date='Jun 23 2003, 01:32 PM

The orginal problem is impossible to solve with only WHOLE numbers. Integers include all REAL numbers, Their NEGATIVES, and zeros.

Originally Posted by Baldy' date='Jun 23 2003, 09:37 AM

[quote name='Joe Flo' date='Jun 23 2003, 01:32 PM'] The orginal problem is impossible to solve with only WHOLE numbers. Integers include all REAL numbers, Their NEGATIVES, and zeros.