Question

Find y as a function of x if y'''-15 y''+56 y^=0 y(0)=5, y'(0)=7, y''(0)=8 y(x)= .

Answer 1

The solution to the differential equation y''' - 15y'' + 56y = 0 is
`y(x) = 4e^4x + 7e^7x + 8e^8x.`

Step-by-step explanation:

The given equation is y''' - 15y'' + 56y = 0.

To find the solution to this differential equation, we assume
`y = e^rx` as a trial solution, where r is a constant.

Substituting this trial solution into the equation gives us the characteristic equation
`r^3 - 15r^2 + 56r = 0.`

Solving this equation using factoring or the quadratic formula, we find r = 4, 7, 8.

Therefore, the general solution to the differential equation is
`y = C1e^4x + C2e^7x + C3e^8x`, where C1, C2, and C3 are constants determined by the initial conditions.

Substituting the initial conditions y(0) = 5, y'(0) = 7, y''(0) = 8 into the general solution, we can find the particular solution.

After substituting the values, the particular solution is
`y(x) = 4e^4x + 7e^7x + 8e^8x.`