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# Find two power series solutions of the given differential equation about the ordinary point x=0. $y^{\prime \prime}-3 x y=0$

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As the given differential equation about the ordinary point $x=0$ , then we'll use the following formula :

$y=\sum_{k=0}^{\infty }\;\;c_{k}\;x^{k}$

Differentiating it with respect to $x$ twice , we obtain

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

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

Substituting in the differential equation we have

$y'' - 3xy=0$

We obtain

$\sum_{k=0}^{\infty }\;(k+1)(k+2)\;c_{k+2}\;x^{k}-3\sum_{k=0}^{\infty }\;\;c_{k}\;x^{k+1} =0$

To obtain $x^{k}$ ,we reduce 1 from any k in this term

$\sum_{k=0}^{\infty }\;(k+1)(k+2)\;c_{k+2}\;x^{k}-3\sum_{k=1}^{\infty }\;\;c_{k-1}\;x^{k} =0$

Substitute in the first term for$k=0$ to form $\sum_{k=1}^{\infty }$ , so

$=2\;c_{2}+\sum_{k=1}^{\infty } \left [ (k+1)(k+2)\;c_{k+2}-3\; c_{k-1} \right ]x^{k}=0$

Since the sum in the last equation must be identically zero then each coefficient of various power of $x$ must be zero in particular ,thus

$c_{2}=0$

$(k+1)\;(k+2)\;c_{k+2}-3\;c_{k-1}=0$

Take the highest $c$ to one side

$c_{k+2}=\dfrac{3}{(k+1)\;(k+2)}\;c_{k-1}$ ,for k = 1,2,3,...

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