Answer to Question #90006 in Differential Equations for lana majeed

Question

Solving this following ODE by Frobenius Method
X^2y’’+(4x-x^2)y’+(2-x)y=0 full solution step-by-step

Answer 1

$$y=\sum_{k=0}^\infty A_k x^{k+r}$$

$$y'=\sum_{k=0}^\infty A_k(k+r) x^{k+r-1}$$

$$y''=\sum_{k=0}^\infty A_k (k+r)(k+r-1)x^{k+r-2}$$

$$x^2\sum_{k=0}^\infty A_k (k+r)(k+r-1)x^{k+r-2}+$$

$$(4x-x^2)\sum_{k=0}^\infty A_k(k+r) x^{k+r-1}+$$

$$(2-x)\sum_{k=0}^\infty A_k x^{k+r}=$$

$$\sum_{k=0}^\infty (A_k (k+r)(k+r-1)+4A_k(k+r)+2A_k)x^{k+r}-$$

$$-\sum_{k=0}^\infty A_k(k+r+1)x^{k+r+1}=$$

$$\sum_{k=0}^\infty A_k((k+r)(k+r-1)+4(k+r)+2)x^{k+r}-$$

$$-\sum_{k=1}^\infty A_{k-1}(k+r)x^{k+r}=$$

$$[(k+r)(k+r-1)+4(k+r)+2=$$

$$k^2+2kr+r^2-k-r+4(k+r)+2=$$

$$(k+r)^2+3(k+r)+2=(k+r+1)(k+r+2) ]$$

$$A_0(r+1)(r+2)x^r+\sum_{k=1}^\infty(A_k(k+r+1)(k+r+2)-A_{k-1}(k+r))x^{k+r}$$

indicial polynom:

$$(r+1)(r+2)=0,r_1=-1,r_2=-2$$

$$A_k(k+r+1)(k+r+2)-A_{k-1}(k+r)=0$$

$$A_k(k+r+1)(k+r+2)=A_{k-1}(k+r)$$ $$case 1: k=1,r=-2$$

$$A_1(1-2+1)(1-2+2)=A_0(1-2)$$

$$A_0=0,A_k=0;$$ $$y = 0$$

$$case 2: k=1,r=-1$$

$$A_1(1-1+1)(1-1+2)=A_0(1-1) = 0$$

$$A_0\not = 0 , A_1=0,A_2=0,...$$

$$y=\frac{A_0}{x}; y = \frac{0}{x}=0, when A_0=0.$$

Answer: y=A₀/x, A₀ is real number.