Math 5 (9.10)

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^3/3+x)y'-y^2x^2/2

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

-Kelsey --Kduncombesmith 20:15, 19 September 2010 (CDT)

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) lololololololololololololololololololol. kthxbai.

--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 , y(2)(0)=y'(0)=0, y(0)=2

Solution: y = 4(x^2)e^(-2x)+4xe^(-2x)+2e^(-2x)

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

Solve: y-4y'+4=0 Given: y(0)=12 y'(0)=-3 Solution: y(t)=12e^2t-27te^2t --Hellison 15:32, 19 September 2010 (CDT)

Solve: 3y[x]+3y'[x]+y[x]=0, y'[0]=4, y[0]=3

Solution: 3e^(-x/2)Cos[x/(2 Sqrt[3])]+11Sqrt[3]E^(-x/2)Sin[x/(2Sqrt[3])]

--Jnaruk 17:45, 19 September 2010 (CDT)

Show that y = C1sin(4x) + C2cos(4x) is a general solution for the differential equation y" + 16y = 0

Soln: Substitute the answer into the original 2nd order differential equation. y' = 4C1cos(4x) - 4C2sin(4x) y" = -16C1sin(4x) - 16C2cos(4x)

Plug in: [-16C1sin(4x) - 16C2cos(4x)] + 16[C1sin(4x) + C2cos(4x)] = 0 = the original differential equation -vchu

Solve: y"'+6y"+9y'=0; y(0)=0; y'(0)=9; y"(0)=36

Solution: y= e^(-3x)[-5cos3x-2sin3x]+5

--Stan 20:21, 19 September 2010 (CDT)

Question: Why is it important to show that a differential equation is exact before solving it?

Answer: It's important to show that a differential equation is exact because of our assumption of the general form we think the solution will take. When we show a differential equation is exact, we show that the derivatives of the partials are the same. Because differentiation is linear (refer to Jade's question), it follows that the solution can be expressed by combining the integrations of each partial. Thus, once we show that the differential equation to be solved is exact, we can automatically jump to the implicit solution F(x,y) = c. --Myuan 00:44, 20 September 2010 (CDT)

Question: Explain why we are *currently* unable to solve a differential equation in the following form: ay+by'+cy=d, where d does not equal zero.

Answer: So, I was a little confused. But basically, we've only learned how to solve homogeneous second order differential equations. Therefore, our solving process will not work for non-homogeneous second order differential equations. Perhaps, we will learn how to solve these later this year, Dr. Raulston?--Cdavis 10:22, 20 September 2010 (CDT)