Hello,
How to solve $4*x^2*y" +12*x*y' +3*y=0, y(4)=\frac18, y'(4)=\frac{-3}{64}$ with the help of Droid48 calculator application? It involves stiff initial value problem as well as vector-value differential equation.
$$4x^2y''+12xy'+3y =0 \quad y(4)=\tfrac18, \, y'(4) = -\tfrac3{64}$$
\begin{align}
&\text{Let} & x &= e^t \\ &&\implies \tfrac{\mathrm dx}{\mathrm dt} &= e^t = x \\[8pt]
&\text{and} & y(t) &= y\big(x(t)\big) \\
&& \tfrac{\mathrm dy}{\mathrm dt} &= \tfrac{\mathrm dy}{\mathrm dx} \tfrac{\mathrm dx}{\mathrm dt} = x\tfrac{\mathrm dy}{\mathrm dx} &\implies \dot y &= xy' \\
&& \tfrac{\mathrm d^2y}{\mathrm dt^2} &= \tfrac{\mathrm dy}{\mathrm dx} \tfrac{\mathrm dx}{\mathrm dt} + x\tfrac{\mathrm d^2y}{\mathrm dx^2} \tfrac{\mathrm dx}{\mathrm dt} \\ &&&= x^2 \tfrac{\mathrm d^2y}{\mathrm dx^2} + x\tfrac{\mathrm dy}{\mathrm dx} &\implies \ddot y &= x^2y'' + xy' \end{align}
And so
\begin{align} 4x^2y''+12xy'+3y &= 4(x^2y''+ xy') + 8xy'+3y \\ &= 4\ddot y + 8\dot y + 3y = 0 \end{align}
The characteristic equation \begin{align} 4r^2 + 8r + 3 &= (2r)^2 + 4(2r) + 3 \\ &= (2r+1)(2r+3) = 0 \end{align}
Has roots \begin{equation}r_1 = -\tfrac12 \quad \text{and} \quad r_2 = -\tfrac32\end{equation}
so \begin{align}y(t) &= c_1 e^{-\frac12 t} + c_2 e^{-\frac32 t} \\ &= c_1 (e^{t})^{-\frac12} + c_2 (e^t)^{-\frac32} \\ y(x) &= c_1 x^{-\frac12} + c_2x^{-\frac32 } \\ &= \frac{c_1}{\sqrt x} + \frac{c_2}{\sqrt{x^3}}\end{align}
And plug in the initial conditions to find the constants.
\begin{align}
&\text{Let} & t &= \ln x \\ &&\implies \tfrac{\mathrm dt}{\mathrm dx}&= \tfrac1x \\[8pt]
&\text{and} & y(x) &= y\big(t(x)\big) \\
&& \tfrac{\mathrm dy}{\mathrm dx} &= \tfrac{\mathrm dy}{\mathrm dt} \tfrac{\mathrm dt}{\mathrm dx} = \tfrac1x\dot y &\implies xy' &= \dot y \\
&& \tfrac{\mathrm d^2y}{\mathrm dx^2} &= -\tfrac1{x^2}\tfrac{\mathrm dy}{\mathrm dt} + \tfrac1{x}(\tfrac{\mathrm d^2y}{\mathrm dt^2}\tfrac{\mathrm dt}{\mathrm dx}) \\ &&&= \tfrac1{x^2}\ddot y - \tfrac1{x^2}\dot y &\implies
x^2y'' &= \ddot y - \dot y\end{align}
And so
\begin{align} 4x^2y''+12xy'+3y &= 4(\ddot y - \dot y) + 12\dot y + 3y \\ &= 4\ddot y + 8\dot y + 3y = 0 \end{align}
And continue as before.
The substitutions $x=e^t$ and $t = \ln x$ are the same, one going one way and one the other. Both are significantly more complicated than just trying $y = x^r$ in the first place, which is the method I would regard as "standard" for an Euler equation.
They are the same, yes. But for me, each is better in parts than the other. They are more complicated than trying $y=x^r$, but that doesn't really help you if the roots of the characteristic equation are not real - unless you were taught the derivation of the real solution (see below). In the case that you have a repeated root, you also need to use the method of reduction of order to get the second solution.
The simplest approach is probably to ask Wolfram Alpha or look for a formula on Wikipedia, but I wouldn't recommend either of those as they tell you nothing about what's actually going on in the equation.
With solutions $$y_1 = x^{u+iv} \qquad y_2 = x^{u-iv}$$
We have \begin{align}
y &= c_1 x^{u+iv} + c_2 x^{u-iv} \\
&= c_1 x^u x^{iv} + c_2 x^u x^{-iv} \\
&= x^u(c_1 x^{iv} + c_2 x^{-iv}) \\
&= x^u(c_1 e^{iv \ln x} + c_2 e^{-iv \ln x}) \\
&= \tfrac12x^u\big((c_1 + c_2) (e^{iv \ln x} + e^{-iv \ln x}) + (c_1 - c_2)(e^{iv \ln x} - e^{-iv \ln x})\big) \\
&= x^u\big(\tfrac{c_1 + c_2}2 \cos{(v \ln x)} + \tfrac{c_1 - c_2}2(\sin{(v \ln x)} \big) \\
&= x^u\big(c_3 \cos{(v \ln x)} + c_4(\sin{(v \ln x)} \big) \end{align}
Not tremendously difficult, but not often taught in my experience.
Oops! I thought it came out a bit easy! I need to sort it out.
Essentially, the real solutions are when $c_1$ and $c_2$ are conjugate pairs, and in this case $c_1 - c_2$ is purely imaginary, so the $i$ cancels. There's also an odd factor of $\tfrac12$ lying around, but that doesn't matter so much.
Something like
With solutions $$ \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} y_1 = x^{u+iv} \qquad y_2 = x^{u-iv}$$
We have \begin{align}
y &= c_1 x^{u+iv} + c_2 x^{u-iv} \\
&= c_1 x^u x^{iv} + c_2 x^u x^{-iv} \\
&= x^u(c_1 x^{iv} + c_2 x^{-iv}) \\
&= x^u(c_1 e^{iv \ln x} + c_2 e^{-iv \ln x}) \\
&= x^u\big((c_1 + c_2) \tfrac12(e^{iv \ln x} + e^{-iv \ln x}) + (c_1 - c_2)\tfrac{i}{2i}(e^{iv \ln x} - e^{-iv \ln x})\big) \\
&= x^u\big((c_1 + c_2)\cos{(v \ln x)} + (c_1 - c_2)i\sin{(v \ln x)} \big)\end{align}
For this expression to be real, we require that $\im{(c_1+c_2)} = 0$ and $\re{(c_1-c_2)} = 0$. This means that we must choose $c_1 = \overline{c_2}$. This also leads to the two coefficients being independent, so we may write.
$$y = x^u\big(c_3\cos{(v \ln x)} + c_4\sin{(v \ln x)} \big) \qquad (c_3,c_4) \in \mathbb R^2$$