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, sign in to answer this question.

Marked as Best Answer

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.

Already have an account? Log in

By signing up, I agree to Wyzant’s terms of use and privacy policy.

Or

To present the tutors that are the best fit for you, we’ll need your ZIP code.

Your Facebook email address is associated with a Wyzant tutor account. Please use a different email address to create a new student account.

Good news! It looks like you already have an account registered with the email address **you provided**.

It looks like this is your first time here. Welcome!

To present the tutors that are the best fit for you, we’ll need your ZIP code.

Please try again, our system had a problem processing your request.