Question

Find the general solution of the given Euler equation on \((0, \infty)\). $$ 3 x^{2} y^{\prime \prime}-x y^{\prime}+y=0 $$

Short answer

The general solution of the given Euler equation on the interval \((0, \infty)\) is: $$ y(x) = x^{\frac{1}{2}}\left[A\cos\left(\frac{\ln{x}}{2}\right) + B\sin\left(\frac{\ln{x}}{2}\right)\right] $$ where A and B are arbitrary constants.

Step-by-step solution

Identify the given Euler equation

The given Euler equation is: $$ 3x^2y'' - xy' + y = 0 $$

Substitute \(x = e^t\)

To transform the Euler equation, let \(x = e^t\). Now, find the derivatives in terms of t: $$ \frac{dx}{dt} = e^t \quad \Rightarrow \quad x' = e^t $$ $$ \frac{d^2x}{dt^2} = e^{t} \quad \Rightarrow \quad x'' = e^t $$ Substitute these derivatives into the original equation, and divide the entire equation by \(x^2 = e^{2t}\): $$ 3y'' - \frac{xy'}{x^2} + \frac{y}{x^2} = 0 $$

Simplify the transformed equation

After dividing, simplify the equation: $$ 3y'' - \frac{e^t}{e^{2t}} + \frac{y}{e^{2t}} = 0 $$ $$ 3y'' - e^{-t}y' + e^{-2t}y = 0 $$

Use the transformation \(y(t) = e^{\alpha t}u(t)\)

Now, we make the substitution \(y(t) = e^{\alpha t}u(t)\). Compute the derivatives and plug them into the transformed equation: $$ y' = e^{\alpha t}u'(t) + \alpha e^{\alpha t}u(t) $$ $$ y'' = e^{\alpha t}u''(t) + 2\alpha e^{\alpha t}u'(t) + \alpha^{2} e^{\alpha t}u(t) $$ Substitute and simplify the equation: $$ 3e^{\alpha t}u''(t) + 3(2\alpha - 1)e^{\alpha t}u'(t) + 3(\alpha^{2}-\alpha)e^{\alpha t}u(t) = 0 $$

Cancel the exponential terms

To cancel the exponential terms, divide the whole equation by \(3e^{\alpha t}\): $$ u'' + (2\alpha - 1)u' + (\alpha^{2} - \alpha)u = 0 $$ To remove the first-order term, we need to select an appropriate α. Choose \(\alpha = \frac{1}{2}\) to eliminate the first-order term: $$ u'' + \frac{1}{4}u = 0 $$

Solve the second-order linear constant-coefficient homogeneous equation

Write the general solution for the transformed equation: $$ u(t) = A\cos\left(\frac{t}{2}\right) + B\sin\left(\frac{t}{2}\right) $$ Now, substitute back the transformation \(y(t) = e^{\frac{1}{2}t}u(t)\): $$ y(t) = e^{\frac{1}{2}t}\left[A\cos\left(\frac{t}{2}\right) + B\sin\left(\frac{t}{2}\right)\right] $$

Convert back to the original variable x

Lastly, substitute \(x = e^t\) to get the general solution in terms of x: $$ y(x) = x^{\frac{1}{2}}\left[A\cos\left(\frac{\ln{x}}{2}\right) + B\sin\left(\frac{\ln{x}}{2}\right)\right] $$ This is the general solution of the given Euler equation on the interval \((0, \infty)\).