Question

If the roots of the characteristic equation of the Euler's ODE has a repeated root m, then what is the correct form of the general solution? (\(c_1,c_2\) are arbitrary constants.)

• A. \(y(x)=c_1x^m+c_2x^{-m}\)
• B. \(y(x)=c_1x^m+c_2x^{-m}\ln x\)
• C. \(y(x)=c_1e^{mx}+c_2e^{-mx}\)
• D. \(y(x)=c_1\cos mx+c_2\sin mx\)

Hint:

Substitute \(x=e^t\) to convert the Euler equation into a constant-coefficient ODE and apply the repeated-root rule.

Answer 1

The Correct Option is B

Step 1: Consider the Euler equation $ax^2y''+bxy'+cy=0$. Trying $y_1=x^m$ gives the indicial equation $am(m-1)+bm+c=0$, i.e. $am^2+(b-a)m+c=0$. A repeated root occurs when $m=\dfrac{a-b}{2a}$.

Step 2: Write the equation in standard form $y''+P(x)y'+Q(x)y=0$ by dividing by $ax^2$: $P(x)=\dfrac{b}{ax}$.

Step 3: Use the reduction-of-order (Abel) formula for the second solution: \[y_2=y_1\int\frac{e^{-\int P\,dx}}{y_1^2}\,dx=x^m\int\frac{x^{-b/a}}{x^{2m}}\,dx\]

Step 4: Since $m$ is the repeated root, $2m=\dfrac{a-b}{a}=1-\dfrac{b}{a}$, so $-\dfrac{b}{a}-2m=-(1-2m)-2m=-1$. The integral becomes \[y_2=x^m\int x^{-1}\,dx=x^m\ln x\]

Step 5: Hence the two independent solutions are $y_1=x^m$ and $y_2=x^m\ln x$, giving the general solution \[y(x)=c_1x^m+c_2x^m\ln x\] which corresponds to option (B). \[\boxed{y(x)=c_1x^m+c_2x^m\ln x}\]