solve the initial value problem. [ y^(3) + y" = x+e^(-x) ] ; y(0)=0 , y''(0)=1?

Question

This is a differential equations problem. I think you have to find y complementary and y particular. The e throws me off. Please show how you did it. Thanks a bunch.

Answer 1

I'm assuming y^(3) refers to y''' rather than the cube of y. In this case, let z = y'' to give the equation
z' + z = x + e^(-x)
The complementary solution for z will be z = ce^(-x), so for particular solution we look for
z = Ax + B + Cxe^(-x)
z' = A + Ce^(-x) - Cxe^(-x)
So we get
Ax + (A+B) + Ce^(-x) = x + e^(-x)
giving A = 1, B = -1, C = 1 and hence z = x - 1 + xe^(-x).
Adding the complementary and particular solutions, we get
z = xe^(-x) + ce^(-x) + x - 1
and since y''(0) = z(0) = c-1 = 1, we get c = 2, so
y''(x) = xe^(-x) + 2e^(-x) + x - 1
By integrating this twice, we can get y(x) which I get to be
y(x) = xe^(-x) + 4e^(-x) + (1/6) x^3 - (1/2) x^2 + ax + b
and from the initial condition y(0) = 0 we get 4+b = 0, so
y(x) = xe^(-x) + 4e^(-x) + (1/6) x^3 - (1/2) x^2 + ax - 4
where a is undetermined, as we'd expect since we have a third-order ODE with only two initial conditions.

If it is indeed supposed to be y^3 + y'' = x + e^(-x), I agree with Dr D that you're not going to get an analytical solution. (We'd normally put the y'' term first when writing this equation, though, and the y^(n) notation is often used to denote the n'th derivative, so I'm optimistic that my solution above is what you wanted.)

Answer 2

As you've written it, I don't think there is a closed form solution. It can be solved numerically using the Euler or Runge Kutta methods. Or maybe you meant y instead of y^3. Or y"'. I don't know.