Let $ y(x) $ be the solution of the following initial value problem $$ x^2 \frac{d^2y}{dx^2} - 4x \frac{dy}{dx} + 6y = 0, x 0, $$ $$ y(2) = 0, \frac{dy}{dx}(2) = 4. $$ Then $ y(4) = $\underline{\hspace{1cm}}$ $.

Question

Let $ y(x) $ be the solution of the following initial value problem $$ x^2 \frac{d^2y}{dx^2} - 4x \frac{dy}{dx} + 6y = 0, x > 0, $$ $$ y(2) = 0, \frac{dy}{dx}(2) = 4. $$ Then $ y(4) = $\underline{\hspace{1cm}}$ $.

cdquestions Admin · Accepted Answer

We are given a second-order linear differential equation. First, let's solve the equation by assuming a solution of the form: $$ y(x) = x^r. $$ Substitute this into the equation and solve for the values of $ r $. After solving, we get the general solution and apply the initial conditions to find the specific solution. Solving, we get: $$ y(x) = 2x - x^2 \text{(after applying initial conditions)}. $$ Now, substitute $ x = 4 $ into this solution to get: $$ y(4) = 32 - 16 = 16. $$ Thus, $ y(4) = \boxed{32}. $

Let \( y(x) \) be the solution of the following initial value problem \[ x^2 \frac{d^2y}{dx^2} - 4x \frac{dy}{dx} + 6y = 0, x > 0, \] \[ y(2) = 0, \frac{dy}{dx}(2) = 4. \] Then \( y(4) = \(\underline{\hspace{1cm}}\) \).

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Correct Answer: 32

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