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Differential equations note 2

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2. Equation reducible to homogeneous linear form The Legendre 's equation is of the form -1 dy d"y+R(a+bx) (a+bx) t P ( a +bx)+P,V)=X (1) where R,R,... ,Pr-l,P, are constants and X is either a constant, or a function ofx alone. This type of equation can be solved by the substitutions a+br=z and =e", whereby the equation reduces to and can be solved. one with constant coeficients 3.3. Ilustrative Examples Example1. Solve: 6x x?d-4+6y= ax d2 [ C.P 1996, 2000] Solution: To solve the equation 4x46y= 6x dr2 lettus substitute logx Then dx =Z, i.e, X=e _dy and 2 d dz (1) 2 dy d dz2 dz with these substitutions equation (1) becomes dyd4+6y =6e33 dz2 dz dz or, dy-5+6y=6e$ For the complementary 2) function , we solve or, dz2 -5+6y=0 If y = e"mz (3) * 0 be a solution of equation (3) , the auxiliary equation is i.e, m=2, m-5m+6 0 3 So the C.F. is ce2 +C2 e37 For the P.I., equation (2) can be written as (D? -5D +6)y=6 where D=- dz 1 P.I. 2 -5D+6 Oe = 6e-6e7 =e (5-5.5+6 Thus the general solution of equation (2) is y = c e27 +C2 e3 e Hence the general solution of the equation (1) is y = cx2+C where c + and c> are two arbitrary constants. Example2. Solve: x [C.P1985,1991,1994,2000 -+y=logx dr Solution: To solve the given equation dy log.x +y= logx *d *y= dx2 let us substitute logx = z, i.e, Then and dx dz ( x = e 2 dz2 dz Under this substitution equation (1) reduces to y=z az or, dy -2+y=z dz (2) CAU complementa ...
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