Question

Use the frobenius method to obtain one solution of the following ode 4xy'' + 2y' + y = 0

Answer 1

Frobenius method for solving a nonhomogeneous linear ordinary differential equation ODE:   4x y'' + 2 y' + y = 0

$y(x)=x^r\ \Sigma_{n=0}^\infty {C_n\ x^n}=\Sigma_{n=0}^\infty {C_n\ x^{n+r}}\\\\y'=\Sigma_{n=0}^\infty {C_n\ (n+r)\ x^{n+r-1}}\\\\y''=\Sigma_{n=0}^\infty {C_n\ (n+r)(n+r-1)\ x^{n+r-2}}\\\\So\ \Sigma_{n=0}^\infty [ 2C_n (n+r)\{2(n+r-1)+1\}]x^{n+r-1}+\Sigma_{n=0}^\infty {C_n\ x^{n+r}}=0\\\\divide\ by\ x^{r}\ and\ equate\ coefficients=0\\\\\Sigma_{n=0}^\infty [ 2C_n (n+r)\{2n+2r-1\}]x^{n-1}+\Sigma_{n=0}^\infty {C_n\ x^n}=0\\\\So\ 2C_0*r*(2r-1)=0\ =\ \textgreater \ r=0\ or\ \frac{1}{2}$

$\\\\For\ n \ge 0,\ C_n=-2\ C_{n+1}(n+r+1)(2n+2r+1)\\\\For\ r=0,\ C_{n+1}=\frac{-C_n}{2(n+1)(2n+1)},\ n \ge 0\\\\For\ r=\frac{1}{2},\ C_{n+1}=\frac{-C_n}{2(n+1)(2n+3)},\ n \ge 0.\\\\Solutions:\ r_1=0:\ y_1(x)=C_0[ 1-\frac{x}{12}+\frac{x^2}{24}-\frac{x^3}{720}.... ]\\\\\ r_2=\frac{1}{2}:\ y_2(x)=C_0[ 1-\frac{x}{6}+\frac{x^2}{120}-\frac{x^3}{5040}.... ]\\\\Finally,\ y(x)=C_1[1-\frac{x}{12}+\frac{x^2}{24}-\frac{x^3}{720}....]+C_2[1-\frac{x}{6}+\frac{x^2}{120}-\frac{x^3}{5040}....]$