Find the solution of 4x^2*y"+8xy'+17y=0, y(1)=2, y'(1)=-3.

Answer: y=2x^(-1/2)*cos(2*lnx)-x^(-1/2)*sin(2*lnx)

y=x^r

y'=rx^r-1

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

r=-1/2±2i

What's the general solution for this?

Find the solution of 4x^2*y"+8xy'+17y=0, y(1)=2, y'(1)=-3.

Answer: y=2x^(-1/2)*cos(2*lnx)-x^(-1/2)*sin(2*lnx)

y=x^r

y'=rx^r-1

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

r=-1/2±2i

What's the general solution for this?

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Hello Sun,

You have actually gotten most of the way there, keep pushing forward. This question is of exactly the same type as a previous one I answered here yesterday, it is an example of an
*Euler equation.*

With the roots r in hand, you can write down the general solution immediately. In this case, since the roots r are complex conjugates, let me denote them by a ± bi, the solution is of the form

y = |x|^{a} ( c_{1 }cos(b ln |x|) + c_{2} sin(b ln |x|) ).

You were given an initial value problem to solve, so you would determine the constants c_{i} by applying the initial conditions, giving

c_{1} = 2 and c_{2} = -1.

Incidentally, it would not be a bad idea to graph the solution and observe how it behaves. You could also learn a lot by changing the initial values and plotting a representative selection of solution curves. Especially note how the solutions behave as x approaches 0, which is the singular point for this equation.

Regards,

Hassan H.