Question

Solve the given initial-value problem. Find the solution to the differential equation y" y' = x, given that y(0) = 3 and y'(0) = 0.

Answer 1

To solve the differential equation y"y' = x with initial conditions y(0) = 3 and y'(0) = 0, we integrate twice and use the initial conditions to find the constants of integration, resulting in the solution y = (1/6)x^3 + 3.

Step-by-step explanation:

To solve the initial-value problem given by the differential equation y"y' = x, with the conditions y(0) = 3 and y'(0) = 0, we will approach the problem in steps.

1. First, observe that we can write the differential equation as y" = x/y'. Since y'(0) = 0, we need to be cautious about dividing by y'.
2. Next, integrate both sides with respect to x to find y'. Introducing a constant of integration C1, we get y' = (1/2)x^2 + C1.
3. Using the initial condition y'(0) = 0 to find C1, we get C1 = 0.
4. So y' = (1/2)x^2. Integrate y' with respect to x to find y. Introducing another constant of integration C2, we get y = (1/6)x^3 + C2.
5. Using the initial condition y(0) = 3 to find C2 gives us C2 = 3.

Therefore, the solution to the initial-value problem is y = (1/6)x^3 + 3.