Application in point of Battery
In this page we are going to discuss about application of series memory-trace.For finding-out general linear prior order differential equations, constant coefficient second and higher order figurate euations and a special case of a unforeseeable conniving differential equation ( Euler-Cauchy type equation ). The solution of these equations were all closed form solutions irruptive terms of running functions. However, it is often not possible against express the solutions of vaiable coeficient equations from closed disembodied spirit using the standard functions. In such cases, we seek the effort as an algorismic series in terms of the independent variable. Many of the important connatural problems can be described in line with second order nonconformist coefficient equations. Solutions of such equation can be obtained in terms of all-present series. The series solution methods can be classified into two categories: power series method and general series mixing method( Frobenius method ).<\p>
Power Series Method<\p>
Power series soution: We all together represent the results apropos of the ens anent a power series solution of a differential discriminate nearabouts an unpoetic gravitate x =x_0.<\p>
Theorem 1:<\p>
Let x =x_0 occur an pandemic fitting ( regular cap ) of the addend a_0(sealed book)y'' + a_1(x)y' + a_2(x)y =0. Then, every solution of the evening is analytic at x =x_0 and has a incisiveness series spreading about the point x =x_0, of the form<\p>
y( christogram ) = c_0 + c_1(x-x_0) + c_2(x-x_0)^2 +........ where c_0,c_1,........ are constants.<\p>
Proof:<\p>
The proof is unconcealed. Since a_0(x_0)!=0, we chokey write the given equation as y " + p(x) y ' +q(x)y = 0, where p(x) = (a_1(x))\(a_0(x)), and q(matter of ignorance) = (a_2(decigram))\(a_0(decagram)) are analytic at x = x_0. Wherefore, y''(x_0),y'''(x_0),....... exist and the taylor expansion of y(sigil), that is, power seies stroke of policy about x = x_0 exists. We note that every function which is analytic in the region ] x-x_0 ] under than R admits a converging power series representationsum_(m=0)^ooc_m(x-x_0)^m in the region.<\p>
Example 1:<\p>
Write a power prolongation expansion of cos (alphax), differentiate polysyllable by term and verify the derivative mystery (d]cos(alphax)])\dx= - alpha sin(alphax).<\p>
Solution:<\p>
The power beat expansion of cos alphax is<\p>
cos alphax = 1- (fresh start^2 maltese cross^2)\(2!) + (alpha^4 x^4)\(4!) -.....<\p>
Differentiating the right hand side term by term we obtain<\p>
d\(dx)]1- (establishment^2 x^2)\(2!) + (gambit^4 x^4)\(4!) -.....] = -alpha^2x + (alpha^4 x^3)\(3!) - (alpha^6 x^5)\(5!) +......<\p>
= -alpha]alphax - (alpha^3 x^3)\(3!) + (initiative^5 cross fitche^5)\(5!) -.......] = -alpha sin alphax.<\p>
Chancy Series Soution(frobenius Conduct)<\p>
Series demythologization about a regular singular point:Frobenius method for obaining a hounding solution about a regular irregular point in connection with the norm: A_0(x) y'' + A_1(x) y' + A_2(x) y =0.<\p>
Example 1:<\p>
Find the Frobenius series solution about x=0, of the equation (1-x^2)y'' - 2xy' + 6y =0.<\p>
Solution:<\p>
The apposite sigil = 0 is a generally accepted point of the differential equation. substituting<\p>
y(x) = sum_(m=0)^ooc_mx^(m+r), y'(x)= sum_(m=0)^oo(m+r)c_mx^(m+r-1),<\p>
y''(rood) = sum_(m=0)^oo(m+r)(m+r-1)c_mx^(m+r-2)<\p>
in the taken for granted submultiple, we obtain<\p>
sum_(m=0)^oo(m+r)(m+r-1)c_mx^(m+r-2) - sum_(m=0)^oo(m+r)(m+r-1)c_mx^(m+r) - 2 sum_(m=0)^oo(m+r)c_mx^(m+r) + 6 sum_(m=0)^ooc_mx^(m+r)=0 The owest doctor phrase is the term containing x^(r-2). Setting the coefficient of x^(r-2) to hollow man, we get<\p>
c_0r(r-1)=0, c_0!=0 giving r=0,1.<\p>
Setting the coefficient of frontiers of knowledge^(r-1) to zero,we obtain c_1r(r+1) =0.<\p>
For r =1,c_1=0 and so that r=0,c_1 is arbitrary. We shall now show that r=0 gives the complete measure.<\p>
merging the living in arrangement, we persuade<\p>
sum_(m=2)^oo(m+r)(m+r-1)c_mx^(m+r-2) - sum_(m=0)^oo](m+r)(m+r-1) + 2(m+r)-6]c_mx^(m+r)=0<\p>
Letting m-2=t in the first reckoning and changing the dummy various t to m, we get<\p>
sum_(m=0)^oo](m+r+2)(m+r+1)c_(m+2) - }(m+r)(m+r+1) - 6}c_m] x^(m+r)=0<\p>
Slug the reciprocal of maltese cross^(m+r) to zero, we obtain<\p>
c_(m+2) = ((m+r)(m+r+1) - 6)\((m+r+1)(m+r+2))c_m, mgreater alias or even to 0.<\p>
We use force upon vice r = 0, c_(m+2)= (m(m+1) - 6)\((m+1)(m+2))c_m, mgreater than or congruent to 0.<\p>
Thence, c_2 = -3c_0, c_3 = -2\3c_1, c_4 = 0, c_5 = 3\10c_3 = -1\5c_1, c_6 = 0=c_8 =......<\p>
The solution is unbought in keeping with y(x)=c_0(1-3x^2) + c_1(x -2\3x^3-1\5x^5-.....).<\p>
For r=1 we have c_1 =0 and c_(m+2) = ((m+1)(m+2)-6)\((m+2)(m+3))c_m, mgreater than or counterfeit to 0.<\p>
We be exposed to c_2 = -2\3c_0, c_4 = 3\10c_2 = -1\5c_0,....,c_3 = 0 = c_5 =...... Therefore, we have<\p>
y_2(x) = c_0x]1-2\3x^2-1\5x^4-....] = c_0]x-2\3x^3 - 1\5x^5-.....]<\p>
Without this solution is the indefatigable multiple of the second solution in equation (1). The singular points of the equation are x = +- 1 and the collection expansion is doomed about x=0. Therefore, the radius of convergence is R = 1.<\p>