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Chapter 8: Problem 6

$$3 x y^{\prime \prime}+(2-x) y^{\prime}-y=0$$

Short Answer

Expert verified
The solution is the general solution of the Cauchy-Euler equation which is \(y = x^{m_1} + x^{m_2}\), where \(m_1\) and \(m_2\) are roots of the quadratic equation obtained after substitution.

Step by step solution

01

Identify the type of the differential equation

The differential equation \(3 x y''+(2-x) y'-y=0\) is a second order non-homogeneous Cauchy-Euler equation.
02

Apply substitution

To solve this equation, one can apply a substitution where \(y = x^m\). Then, use the chain rule to find \(y'\) and \(y''\) and substitute these into the original equation.
03

Solve for m

After substituting \(y\), \(y'\), and \(y''\) into the equation, the differential equation will transform into a quadratic equation in terms of \(m\). Solve the quadratic equation to get the two roots, \(m_1\) and \(m_2\).
04

Form the general solution

The general solution of a Cauchy-Euler equation is \(y = x^{m_1} + x^{m_2}\). Substitute the values of \(m_1\) and \(m_2\) from Step 3 into this equation to form the general solution.

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Most popular questions from this chapter

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