y" - 5 y' + 6 y = x?

Answer 1

what's the question?

if its the GS,i'll try to fiddle out an answer

this is an inhomogeneous DE -the associated homogeneous DE
is
y"-5y'+6y = 0

the auxiliary equation is L^2-5L+6=0

giving L1=2 and L2=3

if the AE has two distinct real roots L1 and L2 the general solution
of the DE is

y=Ce^(L1x) + De^(L2x) where C and D are constants

so general solution is y=Ce^(2x)+De^(3x)

we now have to find the particular integral

as x is linear,we try a solution of the form yp=rx+s

where r and s are coefficients to be determined so that the DE is satisfied-to try this solution we need the first and second derivatives of y

y'=r, y"=0

substituting these into the LHS of the DE gives

0-5r+6(rx+s)= 6rx +6s-5r

therefore,for yp= rx+s to be a solution of the DE we require that

6rx+6s -5r = x comparing coefficients, 6r =1 >>>>> r =1/6

6s-5r =0 >>>> 6s=5/6 >>>>> s=5/36

so, the particular integral yp = x/6+5/36

check by substituting into equation y"-5y' +6y =x

0-5/6+6(x/6+5/36) = -5/6+x +30/36 = x as required

therefore the GS of the inhomogeneous DE y"-5y' +6y=x is

y=Ce^(2x)+De^(3x)+x/6+5/36 where C and D are arbitrary (real) constants

Answer 2

This second order differential equation can be solved by simple methods since it is linear. We first need to find the complementary function, which is the solution to try the solution to the equation y" - 5 y' + 6 y=0. This is because the equation is linear and when this solution is substituted into the equation as part of a sum of other solutions, it will give 0. Hence will not affect the other parts of the sum that will give a non zero result. To find this solution we try the solution y=Aexp(mx) where A is an arbitrary constant. Substituting this into the equation gives
Aexp(mx)(m^2-5m+6)=0. From this we can see that m^2-5m+6=0 or (m-2)(m-3)=0 which gives m=2 or m=3. This gives two possible solutions of the form y=Aexp(mx) and since this is the case it is possible that each one has a different arbitrary coefficient and so the complementary function is y=Aexp(2x)+Bexp(3x) where A and B are arbitrary constants that are set by the boundary conditions.

Next we need to find the particular integral, which is the solution to the differential equation y" - 5 y' + 6 y = x, which when substituted back in gives the x term. Since the term on the right of the equation is a function of x including x^1 terms only we assume a solution of the form y=ax+b. This time a and b are constants to be found since they must leave an exact term "x" when the solution is substituted back in. If y=ax+b then y'=a and y''=0. Putting this into the equation gives (0)-5(a)+6(ax+b)=x or 6ax+6b-5a=x. Comparing the coefficients of x^0 and x^1 we see that 6a=1 and b-5a=0. This means that a=1/6 and 6b=5/6 or b=5/36. The particular integral is therefore y=x/6+5/36.

Since the equation is linear, the solution is the sum of the complementary function and particular integral and is hence:

y=Aexp(2x)+Bexp(3x)+x/6+5/36.

Please note that I have used exp(x) to represent e^x. This is a common expression for this function however if you are not used to it please go back and look at my workings using the notation y=Ae^mx as the trial solution for the complementary function and you will see where the working comes from.

Jez

Answer 3

Aux Eq: M^2-5M+6y=0
(M-2)(M-3)=0
this implies Comp. Function is y=Ae^3x+Be^2x

Particular Integral:
try y=Cx+D
y'=C
y''=0

this implies -5C+6Cx+6D=x
this implies C=1/6
this implies D=5/36
PI: y=x/6+1/30
therefore: GS y=Ae^3x+Be^2x+x/6+5/36

Answer 4

x=y(y-2)(y-3)

Answer 5

The Maid wants a question, answer or there will be arrows whizzing over and into your ugly fizzhog.

Answer 6

Alright smarty pants not everyone has a maths degree. Well actually I do but that's not the point.

Answer 7

x=ur empty head

Answer 8

yes indeed

Answer 9

whats the question???

Answer 10

esteeee, esteee, si, no? o que?