Question

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I AM CURRENTLY IN DIFFERENTIAL EQUATIONS

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1 (a) Determine the largest x-interval where the initial value problem has a unique solution: 1 (22 – 4x)y(6) + Fy(5) + (x + 1)y" + e*y' + (tan x)y In (1 - 1) x2 - 9 with y(2.5) = A, y'(2.5) = B, y"(2.5) C, y" (2.5) = D, y(4) (2.5) E, y(5) (2.5) = F, where A, B, C, D, E, and F are some known constants. (b) Determine whether the set of functions {5, sin’ x, cos 2x} could form a fundamental set of solution of a linear differential equation.

Answer 1

(a)First, in order to use the theorem to find the interval of validity we must write the differential equation in the proper form given in the theorem. So we will need to divide out by the coefficient of the derivative.

y⁽⁶⁾ +y⁽⁵⁾/[ln(x-1)].(x²-4x) +(x+1)y'''/(x²-4x)+e^xy'/(x²-4x) +(tanx)y/(x²-4x) = 1/(x²-4x)(x²-9)

Next, we need to identify where the functions are not continuous. This will allow us to find all possible intervals of validity for the differential equation. So,

So,(x²-4x) will be discontinuous at x=0,4 and (x²-9) at x=3,-3 since these points will give a division by zero. Likewise, ln(x-1) will also be discontinuous at x=1and x=2 since at this point we will have the natural logarithm of zero.

so −∞<x<∞ , exclude x=-3,0,3,1,2,4

and also for ln function x-1>0, x>1 ,1<x<2, 2<x<−∞

−∞ <x<-3 ,-3<x<0 ,0<x<1,1<x<2, 2<x<3, 3<x<4, 4<x<∞ ;

The endpoints of each of the intervals are points where at least one of the two functions is discontinuous. This will guarantee that both functions are continuous everywhere in each interval,

et's identify the actual interval of validity for the initial value problem. The actual interval of validity is the interval that will contain xo=2.5 So, the interval of validity for the initial value problem is.

ans 2<x<3

(B)set of fuctionsf= {5,sin²x,cos2x}

suppose that y1and y2 are two solutions and that W(y1,y2)≠0. Then the two solutions are called a fundamental set of solutions

here f ={f1,f2,f3}

$W(f1,f2,f3)=\begin{vmatrix} f1 & f2 & f3\\ f1' & f2' &f3' \\ f1'' &f2'' & f3'' \end{vmatrix}\\\ =\begin{vmatrix} 5 & sin^{2}x & cos2x\\ 0 & sin2x &-2sin2x \\ 0&2cos2x & -4cos2x \end{vmatrix}$

5(-2sin4x +2sin4x)=0

proved it is not

fundamental set of solutions