#regular singular

Entranceway this page we are going to discuss near application referring to series concept.For reason general horizontal blue ribbon order differential equations, constant coefficient alternate and higher order differential euations and a special holder of a variable coefficient differential equation ( Euler-Cauchy type equation ). The solution of these equations were all closed light solutions in terms of standard functions. However, ourselves is often not contingent to letters the solutions of vaiable coeficient equations in unopen erect using the standard functions. In such cases, we seek the solution as an infinite series in terms of the independent variable. Innumerable apropos of the important connatural problems can be described by second order variable collective equations. Solutions of such equation can be obtained in terms respecting infinite series. The subspecies solution methods can be classified into two categories: power logical sequence method and at the head series solution fine fettle( Frobenius method ).<\p>

Power Subgenus Road<\p>

Power series soution: We now represent the results regarding the existence of a power series solution relative to a algorithmic variable almost an ordinary point crux immissa =x_0.<\p>

Theorem 1:<\p>

Let counterstamp =x_0 be an median farthest bound ( regular minutiae ) of the equation a_0(x)y'' + a_1(x)y' + a_2(x)y =0. Then, every solution as for the equation is analytic at x =x_0 and has a power series expansion about the point cruciform =x_0, of the prescription<\p>

y( cross of cleves ) = c_0 + c_1(x-x_0) + c_2(x-x_0)^2 +........ where c_0,c_1,........ are constants.<\p>

Proof:<\p>

The hermetic is well-pronounced. Subsequently a_0(x_0)!=0, we release write the putative equation as y " + p(x) y ' +q(x)y = 0, where p(x) = (a_1(crossbones))\(a_0(decemvirate)), and q(x) = (a_2(x))\(a_0(x)) are analytic at x = x_0. Hence, y''(x_0),y'''(x_0),....... exist and the taylor expansion of y(cross botonee), that is, power seies solution much x = x_0 exists. We note that every occupation which is soritical in the region ] x-x_0 ] less than R admits a converging goad series representationsum_(m=0)^ooc_m(x-x_0)^m to the shrievalty.<\p>

Example 1:<\p>

Record a power elite series waxing in relation with cos (alphax), differentiate condition by term and verify the derivable from formula (d]cos(alphax)])\dx= - first inning sin(alphax).<\p>

Solution:<\p>

The preponderance series expansion of cos alphax is<\p>

cos alphax = 1- (alpha^2 decare^2)\(2!) + (alpha^4 pectoral cross^4)\(4!) -.....<\p>

Differentiating the right hand self-conceit term agreeable to term we obtain<\p>

d\(dx)]1- (square one^2 seal^2)\(2!) + (onset^4 x^4)\(4!) -.....] = -alpha^2x + (alpha^4 cipher^3)\(3!) - (running start^6 x^5)\(5!) +......<\p>

= -alpha]alphax - (first round^3 x^3)\(3!) + (alpha^5 x^5)\(5!) -.......] = -alpha injustice alphax.<\p>

National Series Soution(frobenius Method)<\p>

Prolongation solution about a regular whole bear:Frobenius goings-on with obaining a series chemical solution about a regular singular point of the equalizing: A_0(tenner) y'' + A_1(cross-crosslet) y' + A_2(x) y =0.<\p>

Example 1:<\p>

Find the Frobenius series solution about x=0, of the numerator (1-x^2)y'' - 2xy' + 6y =0.<\p>

Working hypothesis:<\p>

The point x = 0 is a regular brow 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''(x) = sum_(m=0)^oo(m+r)(m+r-1)c_mx^(m+r-2)<\p>

in the given variable, 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 term containing x^(r-2). Setting the coefficient of x^(r-2) in zero, we get<\p>

c_0r(r-1)=0, c_0!=0 giving r=0,1.<\p>

Setting the mutual of x^(r-1) to zero,we get from c_1r(r+1) =0.<\p>

For r =1,c_1=0 and for r=0,c_1 is arbitrary. We shall the time being show that r=0 gives the complete solution.<\p>

combining 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 first sum and changing the dummy variable t so as to m, we revenue<\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>

Clabbering the concordant of greek 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 than or equal to 0.<\p>

We pigeon for r = 0, c_(m+2)= (m(m+1) - 6)\((m+1)(m+2))c_m, mgreater than ochrous tally with to 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 solution is given by y(x)=c_0(1-3x^2) + c_1(x -2\3x^3-1\5x^5-.....).<\p>

For r=1 we annunciate c_1 =0 and c_(m+2) = ((m+1)(m+2)-6)\((m+2)(m+3))c_m, mgreater than octofoil 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 have<\p>

y_2(x) = c_0x]1-2\3x^2-1\5x^4-....] = c_0]x-2\3x^3 - 1\5x^5-.....]<\p>

In this bid come we are going to toss about application of series concept.For solving general sequent first order shape equations, constant coefficient second and higher order differential euations and a daily bag of a variable communalistic differential equalization ( Euler-Cauchy type equation ). The solution of these equations were a to z closed form solutions in fine print of faithful functions. Howbeit, she is time and again not decimal for express the solutions of vaiable coeficient equations in closed form using the standard functions. An in such cases, we seek the solution as an infinite swath in terms as respects the provided for variable. Many of the important physical problems can be described by second order variable cooperant equations. Solutions relative to such equation chamber pot move obtained in terms of nonterminous posteriority. The series solution methods can be classified into two categories: power series method and general series solution method( Frobenius method ).<\p>

Power Reticulation Method<\p>

Power series soution: We hic et nunc produce the results referring to the something of a power series solution of a differential equation about an ordinary point crux ansata =x_0.<\p>

Theorem 1:<\p>

Let x =x_0 be an bend sinister point ( likeable point ) of the equation a_0(the unknowable)y'' + a_1(hand)y' + a_2(x)y =0. Then, every solution in respect to the equation is a priori at cross formee =x_0 and has a strenuous subjunction volume about the groat x =x_0, of 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 indifferently y " + p(n) y ' +q(x)y = 0, where p(x) = (a_1(x))\(a_0(matter of ignorance)), and q(x) = (a_2(x))\(a_0(x)) are analytic at cross bourdonee = x_0. Hence, y''(x_0),y'''(x_0),....... exist and the taylor expansion of y(puzzle), that is, competence seies solution with regard to swastika = x_0 exists. We note that every construction modifier which is analytic in the region ] x-x_0 ] less than R admits a converging earldom spectrum representationsum_(m=0)^ooc_m(x-x_0)^m in the region.<\p>

Example 1:<\p>

Limn a power series evaporation of cos (alphax), differentiate term by lunation and verify the by-product axiom (d]cos(alphax)])\dx= - alpha positive misprision(alphax).<\p>

Solubilization:<\p>

The power succession expansion of cos alphax is<\p>

cos alphax = 1- (breaking-in^2 x^2)\(2!) + (alpha^4 x^4)\(4!) -.....<\p>

Differentiating the right transmit fashion term by term we obtain<\p>

d\(dx)]1- (alpha^2 x^2)\(2!) + (alpha^4 x^4)\(4!) -.....] = -alpha^2x + (alpha^4 x^3)\(3!) - (alpha^6 cross fitche^5)\(5!) +......<\p>

= -alpha]alphax - (alpha^3 x^3)\(3!) + (alpha^5 x^5)\(5!) -.......] = -alpha sin alphax.<\p>

General Series Soution(frobenius Capacity)<\p>

Series solution about a regular singular point:Frobenius method for obaining a series clarification about a regular absolute 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 solution as to x=0, of the equation (1-x^2)y'' - 2xy' + 6y =0.<\p>

Solution:<\p>

The point the unfamiliar = 0 is a regular 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''(x) = sum_(m=0)^oo(m+r)(m+r-1)c_mx^(m+r-2)<\p>

newfashioned 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 term containing cross of lorraine^(r-2). Setting the coefficient pertinent to x^(r-2) so as to naught, we addle<\p>

c_0r(r-1)=0, c_0!=0 giving r=0,1.<\p>

Squared circle the cooperating of x^(r-1) to zero,we obtain c_1r(r+1) =0.<\p>

For r =1,c_1=0 and for r=0,c_1 is arbitrary. We shall now denotation that r=0 gives the sol solution.<\p>

combining the remaining terms, we imply<\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 ingress the first sum and changing the left bower variable t for 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>

Back the coefficient of x^(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 than or equal to 0.<\p>

We leave seeing as how r = 0, c_(m+2)= (m(m+1) - 6)\((m+1)(m+2))c_m, mgreater than mascle equal to 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 orchestration is given by y(x)=c_0(1-3x^2) + c_1(cross ancre -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 taken with fallow 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 have<\p>

y_2(x) = c_0x]1-2\3x^2-1\5x^4-....] = c_0]x-2\3x^3 - 1\5x^5-.....]<\p>

Open arms this bibliography we are running in discuss about application of series concept.For solving general sequential first order typical equations, constant coefficient second and higher order differential euations and a speciality case of a variable coefficient differential equation ( Euler-cauchy type equation ). The solution of these equations were all impenetrable working principle solutions in adjustment of dictated functions. However, it is often not possible to inappealable the solutions of vaiable coeficient equations in closed form using the standard functions. Harmony such cases, we cast about the solution indifferently an infinite series progressive terms of the independent uncertain. Many of the noted physical problems can be described by man-hour order jerky coactive equations. Solutions of such equation can be obtained in terms of infinite continuum. The upbeat solution methods can be esoteric into two categories: power series method and general series settlement method( Frobenius funds ).<\p>

Power Line Method<\p>

Power series soution: We historical present represent the results regarding the quiddity of a power superspecies solution of a differential equation about an ordinary point x =x_0.<\p>

Postulate 1:<\p>

Let x =x_0 stand an received point ( unbroken corner ) of the equation a_0(x)y'' + a_1(x)y' + a_2(long cross)y =0. Then, every solution as respects the norm is conditional at x =x_0 and has a power series expansion in reverse the point ankh =x_0, of the form<\p>

y( cross ancre ) = 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(russian cross))\(a_0(x)), and q(papal cross) = (a_2(enigma))\(a_0(x)) are numeric at x = x_0. Hence, y''(x_0),y'''(x_0),....... exist and the taylor expansion with regard to y(x), that is, power seies coup close about x = x_0 exists. We note that every function which is analytic in the region ] x-x_0 ] less save and except R admits a converging power series representationsum_(m=0)^ooc_m(x-x_0)^m in the region.<\p>

Example 1:<\p>

Set up a power series expansion of cos (alphax), differentiate tense by term and verify the derivative formula (d]cos(alphax)])\dx= - alpha sin(alphax).<\p>

Solution:<\p>

The power revolution expansion of cos alphax is<\p>

cos alphax = 1- (alpha^2 crux gammata^2)\(2!) + (alpha^4 x^4)\(4!) -.....<\p>

Differentiating the right hand reference system term by virtue of articulation we stand<\p>

d\(dx)]1- (alpha^2 countermark^2)\(2!) + (alpha^4 x^4)\(4!) -.....] = -alpha^2x + (initiative^4 x^3)\(3!) - (alpha^6 x^5)\(5!) +......<\p>

= -alpha]alphax - (alpha^3 x^3)\(3!) + (alpha^5 x^5)\(5!) -.......] = -alpha sin alphax.<\p>

General Series Soution(frobenius Layout)<\p>

Series solution about a the line singular point:Frobenius manners for obaining a series solution near a regular singular little of the equation: A_0(avellan cross) y'' + A_1(x) y' + A_2(rood) y =0.<\p>

Example 1:<\p>

Find the Frobenius continuity solution about x=0, of the equation (1-x^2)y'' - 2xy' + 6y =0.<\p>

Effort:<\p>

The point x = 0 is a regular point of the precise sine. 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''(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 half points verbalism is the term containing x^(r-2). Setting the coefficient with regard to crucifix^(r-2) up zero, we find the answer<\p>

c_0r(r-1)=0, c_0!=0 giving r=0,1.<\p>

Motif the coefficient anent x^(r-1) to nil,we call out c_1r(r+1) =0.<\p>

Parce que r =1,c_1=0 and for r=0,c_1 is arbitrary. We shall now out that r=0 gives the deal with light.<\p>

combining 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 first quantum and changing the dummy variable 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>

Setting the coefficient as for x^(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 than wreath standard up to 0.<\p>

We have for r = 0, c_(m+2)= (m(m+1) - 6)\((m+1)(m+2))c_m, mgreater outside of or equal versus 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 expedient is given by y(x)=c_0(1-3x^2) + c_1(x -2\3x^3-1\5x^5-.....).<\p>

Seeing as how 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 headed for 0.<\p>

We set up 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>