## Related questions with answers

(a)Find a second-order homogeneous linear ODE for which the given functions are solutions. (b)Show linear independence by the Wronskian. (c)Solve the initial value problem.

$x^{m_1}\text{, }x^{m_2}\text{, }y(1)=-2\text{, }y'(1)=2m_1-4m_2$

Solution

Verified#### $\color{#4257b2}{a)}$

Since the given functions are solutions of the Euler-Cauchy equation

$x^2y'' + axy' + by = 0$

we obtain the roots of its auxiliary equation

$\lambda ^2 + (a-1)\lambda + b = 0$

$\lambda _1 = m_1 \quad \wedge \quad \lambda _2 = m_2$

The equation

$(\lambda - m_1)(\lambda - m_2) = 0 \iff \lambda ^2 - (m_1 + m_2)\lambda + m_1 \cdot m_2 = 0$

must be equal to

$\lambda ^2 + (a-1)\lambda + b = 0$

$\begin{align*} & \implies a-1 = -m_1 - m_2 \quad \wedge \quad b = m_1m_2 \\ & \implies a= 1-m_1 - m_2 \quad \wedge \quad b = m_1m_2 \end{align*}$

So, we obtain the ODE:

${\color{#4257b2}{x^2y'' + (1- m_1 - m_2)xy' + m_1m_2y = 0}}$

#### $\color{#4257b2}{b)}$

Let

$y_1 = x^{m_1} \quad \wedge \quad y_2 = x^{m_2}.$

Let's calculate the Wronskian of this two functions.

$\begin{align*} {\color{#c34632}{W(y_1, y_2) }} & \overset{\textcolor{#c34632}{def.}}{=} {\color{#c34632}{\begin{vmatrix} y_1 & y_2 \\ y'_1 & y'_2 \end{vmatrix} }} \\ &= y_1 y'_2 - y_2y'_1 \\ & = x^{m_1} \cdot (x^{m_2})' - x^{m_2} \cdot (x^{m_1})' \\ & = x^{m_1} \cdot m_2 x^{m_2 - 1} + x^{m_2} m_1 x^{m_1 - 1}\\ & = m_2 x^{m_1+ m_2 -1} + m_1 x^{m_1+ m_2 -1} \\ & = x^{m_1+ m_2 -1} (m_2 - m_1) \neq 0 \quad \text{for} \; m_2 \neq m_1 \end{align*}$

$\implies$ $y_1$ and $y_2$ are $\text{\color{#4257b2}{linearly independent}}$.

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