Solution of the differential equation
\[
10x^2\frac{d^2y}{dx^2}-20x\frac{dy}{dx}+22.4y=0
\]
is
\[
y=c_1x^{m_1}+c_2x^{m_2}
\]
where \(m_1 \neq m_2\).
The value of \(m_1+m_2\) (answer in integer) is ______.
Show Hint
Substitute \(y=x^m\) in the Cauchy-Euler equation to get the auxiliary equation \(10m^2-30m+22.4=0\); the sum of roots is \(-b/a\).
Step 1: Use the standard Euler substitution $x=e^t$.
A Cauchy-Euler equation turns into a constant coefficient equation if we set $x=e^t$, so $t=\ln x$. With the operator $D=\dfrac{d}{dt}$, the standard results are
\[ x\frac{dy}{dx}=Dy, \qquad x^2\frac{d^2y}{dx^2}=D(D-1)y \]
Step 2: Rewrite the given equation in terms of $D$.
The equation $10x^2y''-20xy'+22.4y=0$ becomes
\[ 10D(D-1)y-20Dy+22.4y=0 \]
Step 4: Write the characteristic equation.
Since $y=e^{mt}=x^m$ solves this constant coefficient equation, put $D=m$:
\[ 10m^2-30m+22.4=0 \]
This is exactly the same auxiliary equation the direct trial method gives, only reached through the $t=\ln x$ route instead of substituting $y=x^m$ straight into the original equation.
Step 5: Read off the sum of roots directly from the coefficients.
For $am^2+bm+c=0$, the sum of roots is $-b/a$. Here $a=10$, $b=-30$, so
\[ m_1+m_2=-\frac{-30}{10}=3 \]
The individual roots work out to $m_1=1.4$ and $m_2=1.6$, same as before, confirming the sum.
Step 6: Note why this route is useful.
The $x=e^t$ substitution is handy whenever the equation has more than one Cauchy-Euler term mixed with other structure, since it turns the whole problem into an ordinary constant coefficient equation that can be solved with the usual characteristic equation method, rather than guessing a power law form each time.
\[ \boxed{m_1+m_2=3} \]
