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Consider the equation $\left(4-4 x^{2}\right) y^{\prime \prime}-8 x y^{\prime}+3 y=0$
Let $y 1(x)$ be the solution of the equation $(1)$ such that $y_{1}(0)=8, y_{1}^{\prime}(0)=0$ . Find the first three nonzero terms of the power series about 0 of $y_{1}(x) .$

Let $y 2(x)$ be the solution of the equation $(1)$ such that $y 2(0)=0, y_{2}^\prime(0)=1$ . Use a Wronskian to show that $\left\{y_{1}, y 2\right\}$ is a fundamental set for the solutions of the equation $(1)$

$4 y^{\prime \prime}-4 x^{2} y^{\prime \prime}-8 x y^{\prime}+3 y=0$

$y=\sum_{n=0}^{\infty} a n x^{n}$
$y^{\prime}=\sum_{n=0}^{\infty} a_{n}(n) x^{n-1}$
$y^{\prime \prime}=\sum_{n=0}^{\infty} a_{n}(n)(n-1)^{n-2}$

$4\left(\sum_{n=0}^{\infty} a_{n}(n)(n-1) x^{n-2}\right)-4 x^{2}\left(\sum_{n=0}^{\infty} a_{n}(n)(n-1) x^{n-2}\right)-8 x \sum_{n=0}^{\infty} a_{n}(n) x^{n-1}+3 \sum_{n=0}^{\infty} a_{n} x^{n}=0$

$4 \sum_{n=0}^{\infty} a_{n+2}(n+2)(n+1) x^{n}+\sum_{n=0}^{\infty} a_{n}[-4(n)(n-1)-8 n+3] x^{n}=0$

$\sum_{n=0}^{\infty}\left[4\left(a_{n+2}(n+2)(n+1)+a_{n}((-4 n)(n-1)-8 n+3)] x^{n}=0\right.\right.$

$4 a_{n+2}(n+2)(n+1)+a_{n}({-4 n})(n-1)-8 n+3=0$

$a_{n+2}=\frac{-a_{n}(-4 n^2-4 n+3)}{4(n+2)(n+1)}$

$=\frac{a n^{2}+4 n-3}{4(n+2)(n+1)}$ an $\quad n \geq 0$

$n=0 \longrightarrow a_{2}=\frac{-3 a_{0}}{4(2)(1)}=\frac{-3}{8} a_{0}=-3$

$n=1 \rightarrow a_{3}=0=a_{1}$

$n=2 \rightarrow a_4=\frac{-21}{16}$

$y=a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}+a_{4} x^{4}+\cdots$

$y=8+0-3 x^{2}+0-\frac{21}{16} x^{4}+\cdots$

$y=8-3 x^{2}-\frac{21}{16} x^{4}+\cdots$

(b) $W\left(y_{1}, y_{2}\right)(0)=\left|\begin{array}{ll}{y_{1}(0)} & {y_{2}(0)} \\ {y_{1}^{\prime}(0)} & {y_{2}^{\prime}(0)}\end{array}\right|=\left|\begin{array}{ll}{8} & {0} \\ {0} & {1}\end{array}\right|=8-0=8\neq 0$

\$\left\{y_{1}, y_{2}\right\}$ is a fundametal set