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?
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.