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Question

# Show that the general solution of$x y ^ { \prime \prime } + 2 y ^ { \prime } + x y = 0$is $y ( x ) = x ^ { - 1 } ( A \cos x + B \sin x )$.

Solution

Verified
Step 1
1 of 3

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

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

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

Therefore,

\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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