eEuler

Xample Solve the following initial value problem.

$\displaystyle x^{2}y^{\prime\prime} + 4xy^{\prime} + 2y = 0, y(1) = y'(1) = 1$

Answer Put $x = e^{t}, y = x^{\lambda}$ . Then the indicial equation is given by

$\displaystyle \lambda(\lambda - 1) + 4 \lambda + 2 = \lambda^2 + 3\lambda + 2 = (\lambda + 1)(\lambda + 2) = 0.$

From this, we have $\lambda = -1, -2$ . Now this indicial equation is the characteristic equation for the next differential equation.

$\displaystyle \frac{d^{2}y}{dt^{2}} + 3\frac{dy}{dt} + 2y = 0$

Thus, the general solution is

$\displaystyle y = c_{1}e^{-t} + c_{2}e^{-2t} = c_{1}x^{-1} + c_{2}x^{-2}$

Now we apply the initial conditions $y(1) = y'(1) =1$

to obtain

$\displaystyle 1$	$\displaystyle =$	$\displaystyle c_{1} + c_{2}$
$\displaystyle 1$	$\displaystyle =$	$\displaystyle -c_{1} - 2c_{2}$

Solving this, we have $c_{1} = 3, c_{2} = -2$ . Therefore,

$\displaystyle y = 3x^{-1} - 2x^{-2}.$