Find the general solution?

Question

Find the general solution of x^2*y"+3xy'+5y=0 that is valid in any interval not including the singular point.

Answer: y=c1*x^-1*cos(2*ln abs(x))+c2*x^-1*sin(2*ln abs(x))

y=x^r

y'=rx^r-1

y"=r(r-1)x^r-2

 

Answer 1

Hello Sun,

The equation you are trying to solve is called the Euler equation.  The start you have made is the correct one---simply keep going with it.  Once you substitute the hypothesized form of the solution y = x^r into the original equation, you will end up with the equation

x^r( (r(r-1) + 3r + 5 ) = 0

which leads to the quadratic equation

r² + 2r + 5 = 0

with solutions

r = -1 ± 2i.

When the roots r are complex (conjugates) a ± bi, the solution will take the form

y = c₁x^acos(b ln x) + c₂x^asin(b ln x)

for x > 0.

This comes from using the identity

x^r = e^{r ln x}

along with the resolution of the complex exponential into sine and cosine.

I hope this clears things up, this is a good equation to know the solution for, not only because it is an interesting example, but also because it is a favorite of instructors on tests!

Regards,

Hassan H.