Difference between revisions of "Math 5 (9.10)"

Revision as of 14:09, 19 September 2010

September test

Solve y(4)-3y(3)+5y(2)-4y'+4y=0

y=(e^2x)(ax^3+bx^2+cx+d)+fcos(x)+gsin(x)

Solve y(2)-5y(1)+6 y = c1e^-3+c2e^-2-Max

Why, in the proof for solving diff. equations with repeated roots, can we assume that the product of f(t) with e^(r1t) is a solution? Answer: We actually can't assume. This method of assuming is called reduction of order and is more like an educated guess of what the answer is that is later verified. --Deanna (crappy answer I know, but it's the best I can come up with)

3) Solve: y= -((yx^3+x)/3)y'-yx^2

Solution: (yx^3)/3 + yx + (x^3y^2)/6 = C

-Kelsey

Why are you able to add the two different solutions to form a general solution when you solve a linear second order differential equation? For example, with the equation y' ' - 4y' + 3y = 0, if you get r = 1 and r = 3, why is your general solution y = Ae^(x) + Be^(3x) rather than either y = Ce^(x) or y = Ce^(x)?

You are able to add the two different solutions to the second order linear differential equation because differentiation is linear. If the both the first and solutions satisfy the differential equation and both equal 0, then the addition of the two solutions when plugged into the differential equation will also equal 0. --Amai 12:47, 17 September 2010 (CDT)

Can you solve any second order differential equations using separation of variables? Try y^(2)-1 = 0, using both methods. Do you get the same answer? Scott--Swise 12:48, 17 September 2010 (CDT)

Prove whether or not the following differential equation is exact or not, and then solve it:

(5x^2y + y)dx + ((5/x)x^3 + x)dy = 0 Solution:

-Yes, the equation is exact.

-F=(5/3)x^3 + x + C

--Jeallen 12:55, 17 September 2010 (CDT)

Solve: 6y(4)-5y(3)+y(2)=0 y=c1e^(x/2)+c2e^(x/3)+c3+c4x

myles--Mlewis 12:58, 17 September 2010 (CDT)

Solve: 3y(2) + 18iy' + 27y = 0 y(0)=5, y'(0)=12

Solution:

y= (1/5) (cos(-4x) - sin(-4x) + 24 cos(-7x/3) - 24sin(-7x/3))

-Christy--Clee 20:44, 18 September 2010 (CDT)

Solve: y+4y=0

Solution: y= C1(cos2x)+C2(sin2x)

--Bkumar 10:43, 19 September 2010 (CDT)

What is the importance of linear independence when solving differential equations (second-order and higher)?

A differential equation must have as many solutions as its highest order. The homogeneous solution is a linear combination of all these solutions. We want each of these solutions to be "distinct" from one another (obviously, otherwise they wouldn't be separate solutions). Linear independence determines if these solutions are distinct from one another. If two functions are linearly independent, then you cannot multiply one function by something to get the other function. So, if the solutions were not linearly independent, than you could technically just multiply one of the solutions by something to get the other (i.e. multiply solution 1 by the constant c1 to get solution 2), and you wouldn't have two distinct solutions to your differential equation.

For example, if we solve the solution to a differential equation to be: y(x) = c1*f(x) + c2*g(x), but g(x) was not linearly independent of f(x), then we could simply multiply g(x) by a constant c (in this case c1) to get f(x) and the solutions would essentially be the same (which doesn't work if we need two distinct solutions or more). Thus, we need to make sure the solutions to a differential equation are linearly independent of each other in order to obtain distinct solutions to the differential equation. --Jjuzswik 13:53, 18 September 2010 (CDT)

Y1=C1*X*e^(5x), Y2=C2*e^(5x) Ensure that these two functions are linearly independent by use of a Wronskian matrix. Then, assuming they are linear, find the initial second order differential equation for which these two functions are general solutions.

Solution: They are linear. y"-10Y'+25Y=0 --Gpeterkin 11:39, 19 September 2010 (CDT)

Solve: y+9y=0; y(0)=5, y'(0)=1

Solution: r=+/-3i y1=C1cos(3x)+C2sin(3x) y2=C3cos(-3x)+C4sin(-3x) cos(-3x)=cos(3x) sin(-3x)=-sin(3x) y=C5cos(3x) + C6sin(3x) y'=-3C5sin(3x) + 3C6Cos(3x) C5= 5; C6=1/3 y=5cos(3x) + (1/3)sin(3x) --Wrogers 11:50, 19 September 2010 (CDT)

Solve: dy/dx = ((-y^3)/3-y/x)/(ln x + xy^2)

Solution: F = (xy^3)/3 + y*ln x + C

--Awinograd 13:02, 19 September 2010 (CDT)

Solve: y(3)+6y(2)+12y'+8y=0

Solution: y = c3(x^2)e^(-2x)+c2xe^(-2x)+c1e^(-2x)

--Jpham 14:08, 19 September 2010 (CDT)