Solution Found!
The differential equation is known as Hermite’s equation
Chapter 6, Problem 51E(choose chapter or problem)
The differential equation
\(y^{\prime \prime}-2 x y^{\prime}+2 \alpha y=0\)
is known as Hermite’s equation of order after the French mathematician Charles Hermite (1822–1901). Show that the general solution of the equation is \(y(x)=c_{0} y_{1}(x)+c_{1} y_{2}(x)\), where
\(y_{1}(x)=1+\sum_{k=1}^{\infty}(-1)^{k} \frac{2^{k} \alpha(\alpha-2) \cdots(\alpha-2 k+2)}{(2 k) !} x^{2 k}\)
\(y_{2}(x)=x+\sum_{k=1}^{\infty}(-1)^{k} \frac{2^{k}(\alpha-1)(\alpha-3) \cdots(\alpha-2 k+1)}{(2 k+1) !} x^{2 k+1}\)
are power series solutions centered at the ordinary point 0.
Text Transcription:
y^prime\prime-2xy^prime+2alphay=0
y(x)=c_0y_1(x)+c_1y_2(x)
y_1(x)=1+sum_k=1^infty(-1)^kfrac2^kalpha(alpha-2)cdots(alpha-2k+2)(2 k)!x^2k
y_2(x)=x+sum_k=1^infty(-1)^kfrac2^k(alpha-1)(alpha-3)cdots(alpha-2k+1)(2k+1)!x^2k+1
Questions & Answers
QUESTION:
The differential equation
\(y^{\prime \prime}-2 x y^{\prime}+2 \alpha y=0\)
is known as Hermite’s equation of order after the French mathematician Charles Hermite (1822–1901). Show that the general solution of the equation is \(y(x)=c_{0} y_{1}(x)+c_{1} y_{2}(x)\), where
\(y_{1}(x)=1+\sum_{k=1}^{\infty}(-1)^{k} \frac{2^{k} \alpha(\alpha-2) \cdots(\alpha-2 k+2)}{(2 k) !} x^{2 k}\)
\(y_{2}(x)=x+\sum_{k=1}^{\infty}(-1)^{k} \frac{2^{k}(\alpha-1)(\alpha-3) \cdots(\alpha-2 k+1)}{(2 k+1) !} x^{2 k+1}\)
are power series solutions centered at the ordinary point 0.
Text Transcription:
y^prime\prime-2xy^prime+2alphay=0
y(x)=c_0y_1(x)+c_1y_2(x)
y_1(x)=1+sum_k=1^infty(-1)^kfrac2^kalpha(alpha-2)cdots(alpha-2k+2)(2 k)!x^2k
y_2(x)=x+sum_k=1^infty(-1)^kfrac2^k(alpha-1)(alpha-3)cdots(alpha-2k+1)(2k+1)!x^2k+1
ANSWER:Step 1 of 6
In this problem we need to show that the general solution of the differential equation is
, where