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Question

Show that the general solution of

xy+2y+xy=0x y ^ { \prime \prime } + 2 y ^ { \prime } + x y = 0

is y(x)=x1(Acosx+Bsinx)y ( x ) = x ^ { - 1 } ( A \cos x + B \sin x ).

Solution

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According to the appropriate theorem we conclude and express given equation:

x2y+2xy+x2y=0x^{2} y^{\prime \prime}+2 x y^{\prime}+x^{2} y=0

A=2B=0C=1q=2\begin{align*} A &=2\\ B &=0\\ C &=1\\ q &=2 \end{align*}

Therefore,

α=1A2=122=12β=q2=22=1k=2Cq=212=1p=(1A)24(B)q=(12))24(0)2=12\begin{align*} \alpha &= \dfrac{1-A}{2} = \dfrac{1-2}{2} = -\dfrac{1}{2}\\ \beta &= \dfrac{q}{2} = \dfrac{2}{2} = 1 \\ k &= \dfrac{2\sqrt{C}}{q} = \dfrac{2 \sqrt{1}}{2} = 1 \\ p &= \dfrac{\sqrt{\left(1-A\right)^{2} - 4\left(B\right)}}{q} = \dfrac{\sqrt{\left(1-2)\right)^{2} - 4\left(0\right)}}{2}= \dfrac{1}{2} \end{align*}

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