## Solve the Differential Equation \(x^{2} y^{\prime \prime}+11 x y^{\prime}+25 y=0 ?\)

Answer 1:

\(y(x)=c_{1} x^{-5}+c_{2} x^{-5} \ln x\)

Explanation:

Substitute the variable:

So that:

\(y^{\prime}(x)=\frac{1}{x} \phi^{\prime}\)

\(y^{\prime \prime}(x)=\frac{1}{x^{2}}\left(\phi^{\prime \prime}-\phi^{\prime}\right)\)

substituting in the original equation:

\(x^{2} y^{\prime \prime}+11 x y^{\prime}+25 y=0\)

\(\phi^{\prime \prime}-\phi^{\prime}+11 \phi^{\prime}+25 \phi=0\)

\(\phi^{\prime \prime}+10 \phi^{\prime}+25 \phi=0\)

This is a second order equation with constant coefficients, so we can solve the characteristic equation:

\(\lambda^{2}+10 \lambda+25=0\)

\((\lambda+5)=0\)

\(\lambda=-5\)

so the general solution is:

\(\phi(t)=c_{1} e^{-5 t}+c_{2} t e^{-5 t}\)and undoing the substitution:

\(y(x)=\phi(\ln x)=c_{1} e^{-5 \ln x}+c_{2} \ln x e^{-5 \ln x}\)

\(y(x)=c_{1} x^{-5}+c_{2} x^{-5} \ln x\)

Answer 2:

\(y=(A \ln x+B) x^{-5}\)

Explanation:

We have:

This is a Euler-Cauchy Equation which is typically solved via a change of variable. Consider the substitution:

\(x=e^{t} \Rightarrow x e^{-t}=1\)Then we have,

\(\frac{d y}{d x}=e^{-t} \frac{d y}{d t}, \text { and, } \frac{d^{2} y}{d x^{2}}=\left(\frac{d^{2} y}{d t^{2}}-\frac{d y}{d t}\right) e^{-2 t}\)Substituting into the initial DE [A] we get:

\(x^{2}\left(\frac{d^{2} y}{d t^{2}}-\frac{d y}{d t}\right) e^{-2 t}+11 x e^{-t} \frac{d y}{d t}+25 y=0\)

\(∴\left(\frac{d^{2} y}{d t^{2}}-\frac{d y}{d t}\right)+11 \frac{d y}{d t}+25 y=0\)

\(∴ \frac{d^{2} y}{d t^{2}}+10 \frac{d y}{d t}+25 y=0 \ldots . . \text { [B] }\)

This is now a second order linear homogeneous Differentiation Equation. The standard approach is to look at the Auxiliary Equation, which is the quadratic equation with the coefficients of the derivatives, i.e.

\(m^{2}+10 m+25=0\)We can solve this quadratic equation, and we get a real repeated root:

\(m=-5\)Thus the Homogeneous equation [B] has the solution:

\(y=(A x+B) e^{-5 t}\)Now we initially used a change of variable:

\(x=e^{t} \Rightarrow t=\ln x\)So restoring this change of variable we get:

\(y=(A \ln x+B) e^{-5 \ln x}\)

\(∴ y=(A \ln x+B) x^{-5}\)

Which is the General Solution.