Question

For the following Euler-Cauchy equation: x2y" + axy + by = 0 a) Show that y(x)-xrnis a solution where mis equal to m -(1-a) | (1-а)2-b b) Show that for the case when ^1 -a)2 - b 0, the general solution is equal to 4. 4 1-a y(x) = x-2-(G + c2 In x) c) Solve the following problem x2y"-5xy' + 9y-0, y(1)-0.2, y'(1)-0.3 d) Show that for the case when-(1-a)2-b 〈 0, the general solution is equal to 1-а y(x) = xT(c1 cos (w In x) + c2 sin (winx)), |b--(1-a)2 w= 4 e) Solve the following problem x2y" +3xy'2y , y(1) 0.2, y(1) 0.3

Answer 1

Solution:

Given that,
" Given that y (x) = x^m is a solution where m is equal to m = 1/2 (1-a) plusminus squareroot 1/4 (1-a)^2-b y (x) = x^n be a solution , then x^2. M (m-1) x^n-2 +ax.ln x^n-1 +bx^n = 0 implies { m (m-1) +am +b }x^m = 0 implies m (m-1) +am +b = 0 implies m^2-m + 1m +b = 0 implies m^2 -m (1-a) +b = 0 implies m = (1-a) plusminus squareroot (1-a)^2 -4b/2 m = 1/2 (1-a) plusminus squareroot 1/4 (1-a)^2 -b hence obtained. \r\n x^2y^""-5xy^'+ 9y = 0, y (1) = 0.2, y^'(1) = 0.3 transformed equation, d^2y/dt^2 -dy/dt +ady/dt +by = 0 here, a = -5, b = 9 then, implies d^2y/dt^2 -dy/dt-5dy/dt + 9y = 0 implies d^2y/dt^2-6 dy/dt + 9y = 0 now, the C.I equation is m^2-6m + 9 = 0 m^2-3m-3m + 9 = 0 m (m-1)-3 (m-3) = 0 (m-3)(m-3) = 0 m = 3, 3 "