Question

just focus on A,B,D

1. Homogeneous ODE Find a general solution of the linear non-constant coefficient, homogeneous ODE for y(x) x3y'" – 3xy" + (6 – x2)xy' – (6 – x?)y = 0 as follows. a) You are given that yı(x) = x is a solution to the above homogeneous ODE. Confirm (by substitution) that this is the case. b) Apply reduction of order to find the remaining two solutions, then state the general solution. (Hint: The substitution y2(x) = y1(x)u(x) should give a constant coefficient ODE.] c) [MATLAB Grader] Use computer algebra to find the Wronskian W(x) of the three solutions to the homogeneous ODE. d) Use the expression for W(x) from part (c) to apply the Wronskian test. For what values of x are the solutions linearly independent? e) [MATLAB Grader] Use dsolve to double check the general solution you found in part (b) is correct.

Answer 1

We have to solve the problem part by part. In the part (a) we are asked to check whether the given function is a solution or not. To check this we have to just put the values on lhs and check whether lhs=rhs.

In the next part (b) we have to reduce the order of the given ode in teems of a known solution and find the other two solutions.

Next for part (d) we just have to find tge wronskian of the three solutions that we found previous and conclude about the values of x for which the solutions will be lonearly independent. The detailed method is given below:

The given ODE is - mºy" - 3u2y" +(6-0 ) ny' - (-4) = (-) (2) first we are asked to prove that the given y, (u) an is a solution to the given ODE (1) s to prove this, we have to just put the given to tune y, (n) = n in the L.H.S of H and check whether that leads to LAS - RHS. LHS no y, - 342 y, 11 & (han2 my,' - (6-42) y, - R.H.S yi' = -1 y," = 0 Y1 = 0 the given y, (n)=n! So, it is proved that is a solution of (i). (6) I'm the reduction of onder method we have to pub 7 (4) = y, (2) us. so that the given equation (3) will be reduced to a 2nd order differential equation in un) and we will be able to find the other two solutions. So, 960) = a (n) will give the complete solution where any will give the other two solutions, Now, g'(u) = xu(a) & ulu) (n) = nu" (n) + Qu'(ng onde y"() = xu" (M) + Bulling Since yeng is the complete solution of (1) , we put all these a in (7) and get 3şnu"(m) + 3 u "(n)}- 342 { nu "(m) + Qu'(W)} & (642) && na '(m) + u(n)} - (6 ~42) num) = 0 on + "(n)-n9 fuling so

om, on, " (w-u' (u) = 0 (: * 40 ) Cu" - u) = Integrating both sides we get u"(a) u (n) - - . (where is integration com tame function we go Taking u(n) = e me as a complementary auxiliary equation as — m2 = 0 So, the complementary formation is - u (n) = cen & Ge - (where e are constants Now, particulas integ Pif = D - - 1 = (-1 (1-02)." ces the geveral solution of (ii) is given by - un) = cen & chennec This the complete solution is given by y (n) = xu(a) 96) = a rent one- On

(d) Now, we have to find the wrongkian of the there solutiony of equation (1) We have got the general nocution of ODE (i) as - y (n) = c non & green - cx where we see that non , ne-a and (w) are the three solutions nen ne neon nentzen I, Y' Y' = ne* (ne-"-26-) - neun (ne" + 2e") - |- - - - - + - - - - - - 2n - 23 & 4n = -2m3 This in the required Wrongkian. Now, the theory of three solutions y, Y2 Wrong kian and y3 states not that if rero, then the Wrongkian af the solutions are is linearly independent So for a mot bening rere gives w to and hence we con state that for every mon- rere value of the solutions will be independens.