Question:
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 ______.
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 ______.
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 \(-b/a\).
Updated On: Jul 28, 2026
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Correct Answer: 3
Solution and Explanation
Step 1: Identify the type of differential equation.
The given equation is
\[ 10x^2\frac{d^2y}{dx^2}-20x\frac{dy}{dx}+22.4y=0 \]
Every term has the power of \(x\) matching the order of the derivative it multiplies. This is a Cauchy-Euler (equidimensional) equation, and it is solved by trying a power law solution \(y=x^m\).
Step 2: Substitute the trial solution.
For \(y=x^m\),
\[ \frac{dy}{dx}=mx^{m-1}, \qquad \frac{d^2y}{dx^2}=m(m-1)x^{m-2} \]
Put these into the equation:
\[ 10x^2\cdot m(m-1)x^{m-2}-20x\cdot mx^{m-1}+22.4x^m=0 \]
\[ 10m(m-1)x^m-20mx^m+22.4x^m=0 \]
Since \(x^m\) is never zero for \(x>0\), divide through by \(x^m\).
Step 3: Form the auxiliary (indicial) equation.
\[ 10m(m-1)-20m+22.4=0 \]
\[ 10m^2-10m-20m+22.4=0 \]
\[ 10m^2-30m+22.4=0 \]
Divide the whole equation by \(10\):
\[ m^2-3m+2.24=0 \]
Step 4: Use the sum of roots rule.
For a quadratic \(am^2+bm+c=0\), the sum of the roots is \(-b/a\), so here
\[ m_1+m_2=-\frac{-3}{1}=3 \]
We do not even need to find \(m_1\) and \(m_2\) separately to answer the question, but it is worth checking them.
Step 5: Solve for the individual roots and verify.
\[ m=\frac{3\pm\sqrt{9-4(2.24)}}{2}=\frac{3\pm\sqrt{0.04}}{2}=\frac{3\pm0.2}{2} \]
This gives
\[ m_1=1.6,\qquad m_2=1.4 \]
Check: \(m_1+m_2=1.6+1.4=3\), which matches Step 4.
Final Answer:
The value of \(m_1+m_2\) is
\[ \boxed{3} \]
The given equation is
\[ 10x^2\frac{d^2y}{dx^2}-20x\frac{dy}{dx}+22.4y=0 \]
Every term has the power of \(x\) matching the order of the derivative it multiplies. This is a Cauchy-Euler (equidimensional) equation, and it is solved by trying a power law solution \(y=x^m\).
Step 2: Substitute the trial solution.
For \(y=x^m\),
\[ \frac{dy}{dx}=mx^{m-1}, \qquad \frac{d^2y}{dx^2}=m(m-1)x^{m-2} \]
Put these into the equation:
\[ 10x^2\cdot m(m-1)x^{m-2}-20x\cdot mx^{m-1}+22.4x^m=0 \]
\[ 10m(m-1)x^m-20mx^m+22.4x^m=0 \]
Since \(x^m\) is never zero for \(x>0\), divide through by \(x^m\).
Step 3: Form the auxiliary (indicial) equation.
\[ 10m(m-1)-20m+22.4=0 \]
\[ 10m^2-10m-20m+22.4=0 \]
\[ 10m^2-30m+22.4=0 \]
Divide the whole equation by \(10\):
\[ m^2-3m+2.24=0 \]
Step 4: Use the sum of roots rule.
For a quadratic \(am^2+bm+c=0\), the sum of the roots is \(-b/a\), so here
\[ m_1+m_2=-\frac{-3}{1}=3 \]
We do not even need to find \(m_1\) and \(m_2\) separately to answer the question, but it is worth checking them.
Step 5: Solve for the individual roots and verify.
\[ m=\frac{3\pm\sqrt{9-4(2.24)}}{2}=\frac{3\pm\sqrt{0.04}}{2}=\frac{3\pm0.2}{2} \]
This gives
\[ m_1=1.6,\qquad m_2=1.4 \]
Check: \(m_1+m_2=1.6+1.4=3\), which matches Step 4.
Final Answer:
The value of \(m_1+m_2\) is
\[ \boxed{3} \]
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