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?

Tutors, please sign in to answer this question.

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.

Emily R.

Expert Regents, AP and College Level Physics and Math Tutor

Long Island City, NY

5.0
(146 ratings)

Avi N.

Math Made Easy & Computer Skills Too

Tenafly, NJ

5.0
(168 ratings)

John Z.

Chemistry, Physics, SAT Math, and Calculus Tutor; CU Chemical Engineer

Brooklyn, NY

5.0
(306 ratings)