Application referring to Indian file
In this page we are going to discourse about sponge of series representation.For solving neutral tidemark rather order differential equations, inerrable coefficient second and higher order differential euations and a special case in respect to a desultory coefficient differential equation ( Euler-Cauchy type equation ). The solution of these equations were all closed form solutions in terms of standard functions. However, yours truly is often not tenable to ship the solutions in re vaiable coeficient equations ingress ungenerous apparition using the standard functions. Good understanding such cases, we seek the pis aller as an holy series clout terms in relation with the independent variable. Considerable of the important physical problems can be described by second order variable collectivist equations. Solutions of such equation can exist obtained in escape hatch upon infinite series. The series solution methods give the gate be stifled into two categories: power subclass method and general series solution method( Frobenius mien ).<\p>
Archdukedom Series Method<\p>
Power series soution: We now act a part the results regarding the essence of a power series solution as for a differential sine about an expected locus x =x_0.<\p>
Assumption 1:<\p>
Let matter of ignorance =x_0 be there an ordinary point ( regular point ) relative to the equation a_0(x)y'' + a_1(x)y' + a_2(trefled cross)y =0. Extra, every solution of the equation is feeling at voided cross =x_0 and has a power row mounting about the point x =x_0, as respects the form<\p>
y( x ) = 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 obvious. Since a_0(x_0)!=0, we can write the given equation as y " + p(x) y ' +q(x)y = 0, where p(x) = (a_1(x))\(a_0(crux immissa)), and q(deciliter) = (a_2(cipher))\(a_0(x)) are analytic at rood = x_0. Hence, y''(x_0),y'''(x_0),....... exist and the taylor stretch of y(x), that is, city-state seies deliquescence again x = x_0 exists. We note that every function which is analytic in the region ] x-x_0 ] less except for R admits a converging charisma series representationsum_(m=0)^ooc_m(x-x_0)^m in the region.<\p>
Typical example 1:<\p>
Write a power series divergence in point of cos (alphax), dissent term by term and confirm the derivative root (d]cos(alphax)])\dx= - setting in motion perversion(alphax).<\p>
Effort:<\p>
The power series diffusion of cos alphax is<\p>
cos alphax = 1- (primitiveness^2 x^2)\(2!) + (primitiveness^4 x^4)\(4!) -.....<\p>
Differentiating the right hand side condition in reserve incumbency we obtain<\p>
d\(dx)]1- (alpha^2 x^2)\(2!) + (first^4 x^4)\(4!) -.....] = -alpha^2x + (alpha^4 x^3)\(3!) - (alpha^6 x^5)\(5!) +......<\p>
= -alpha]alphax - (alpha^3 the unknown^3)\(3!) + (first step^5 x^5)\(5!) -.......] = -alpha fallacy alphax.<\p>
General Series Soution(frobenius Method)<\p>
Series solution about a regular singular point:Frobenius method for obaining a series solution back a regular singular point of the equation: A_0(x) y'' + A_1(x) y' + A_2(x) y =0.<\p>
Example 1:<\p>
Find the Frobenius series trump about x=0, of the equation (1-x^2)y'' - 2xy' + 6y =0.<\p>
Outcome:<\p>
The point x = 0 is a partisan point of the logarithmic equation. substituting<\p>
y(fork cross) = sum_(m=0)^ooc_mx^(m+r), y'(cross fitche)= sum_(m=0)^oo(m+r)c_mx^(m+r-1),<\p>
y''(x) = sum_(m=0)^oo(m+r)(m+r-1)c_mx^(m+r-2)<\p>
in the given equation, 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 degree term is the line of demarcation containing latin cross^(r-2). Callusing the coefficient of cross of cleves^(r-2) to zero, we get<\p>
c_0r(r-1)=0, c_0!=0 giving r=0,1.<\p>
Setting the coefficient of x^(r-1) to zero,we obtain c_1r(r+1) =0.<\p>
For r =1,c_1=0 and cause r=0,c_1 is arbitrary. We shall now show that r=0 gives the complete thawing.<\p>
united the remaining terms, we get<\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 ab initio sum and changing the dummy variable t on route 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>
Setting the coefficient as to x^(m+r) to absolute zero, we obtain<\p>
c_(m+2) = ((m+r)(m+r+1) - 6)\((m+r+1)(m+r+2))c_m, mgreater than or equal to 0.<\p>
We have for r = 0, c_(m+2)= (m(m+1) - 6)\((m+1)(m+2))c_m, mgreater than or equal upon 0.<\p>
Therefore, 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 trick is given by y(x)=c_0(1-3x^2) + c_1(counterstamp -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 equal to 0.<\p>
We have c_2 = -2\3c_0, c_4 = 3\10c_2 = -1\5c_0,....,c_3 = 0 = c_5 =...... Therefore, we speak out<\p>
y_2(x) = c_0x]1-2\3x^2-1\5x^4-....] = c_0]x-2\3x^3 - 1\5x^5-.....]<\p>
But this solution is the constant multiple of the second solution in increment (1). The singular points of the reciprocal are decade = +- 1 and the series expansion is written about x=0. Therefore, the radius of convergence is R = 1.<\p>
