Convert IVP to an equivalent Volterra integral equation Convert the following initial value problem to an equivalent Volterra integral equation:

$  
  \begin{cases}
    u'' -u' \sin x + \Bbb e ^x u= x   \\
       u(0)=1\\
   u'(0)=-1\\
  \end{cases}
$

I have solved it and I have got the solution

$$ x- \sin x + x \Bbb e ^x - \Bbb e ^x - \int\limits _{0} ^ x u(t) \Big(\Bbb e ^x (x-t) - \sin x \Big) \Bbb d t ,$$

but our instructor in the university has solved it using another method and he has obtained the solution

$$\frac {x^3} 3 -x +1 + \int \limits _{0} ^x \Big( \sin t -(x-t)(\cos t + \Bbb e ^t) \Big) u(t) \Bbb d t .$$

My question is: why are the two solutions different? Is my solution wrong?

Thanks for help.
As written above, both solutions are wrong. To verify your solution, derive it once; you will get $u'(x) = 1 - \cos x + x \Bbb e ^x + u(x) \sin x$. Deriving it a second time: $u'' (x) = \sin x + \Bbb e ^x + x \Bbb e ^x + u'(x) \sin x - u(x) \cos x$. This clearly does not look like your original equation.

Concerning your instructor's solution, deriving it once gives $u'(x) = x^2 - 1 + u(x) \sin x$; deriving once more gives $u'' (x) = 2x - u(x) \cos x + u' (x) \sin x$. Again, this is very far from the given equation.

Maybe you have made some typing mistakes when writing the above formulae, or maybe you have not taken notes correctly. As you are going to see next, your instructor's formula is very close to the correct one.

Let us start by integrating the given equation; we get

$$u'(y) - u'(0) = \int \limits _0 ^y u' (t) \sin t \Bbb d t - \int \limits _0 ^y \Bbb e ^t u(t) + \int \limits _0 ^y t \Bbb d t .$$

Using that $u'(0) = -1$ and integration by parts in the first integral we get

$$u'(y) = u(y) \sin y - \int \limits _0 ^y u(t) (-\cos t) \Bbb d t - \int \limits _0 ^y \Bbb e ^t u(t) + \frac {y^2} 2 -1 ,$$

or equivalently

$$u'(y) = \frac {y^2} 2 -1 + u(y) \sin y + \int \limits _0 ^y (\cos t - \Bbb e ^t) u(t) \Bbb d t .$$

Now, let us integrate this once more:

$$u(x) - u(0) = \int \limits _0 ^x \frac {y^2} 2 -1 \Bbb d y + \int \limits _0 ^x u(y) \sin y \Bbb d y + \int \limits _0 ^x \int \limits _0 ^y (\cos t - \Bbb e ^t) u(t) \Bbb d t \Bbb d y .$$

We shall now use $u(0) = 1$ and we shall change the order of integration in the last integral in order to get

$$u(x) = \frac {x^3} 6 - x + 1 + \int \limits _0 ^x u(y) \sin y \Bbb d y + \int \limits _0 ^x \int \limits _t ^x (\cos t - \Bbb e ^t) u(t) \Bbb d y \Bbb d t .$$

Since the function in the last integral does not depend on $y$, if we change the letter from $y$ to $t$ in the first integral we get

$$u(x) = \frac {x^3} 6 - x + 1 + \int \limits _0 ^x \Big( \sin t + (x-t) (\cos t - \Bbb e ^t) \Big)u(t) \Bbb d t .$$

As you can see, the only differences between the instructor's formula and the correct one are the denominator ($6$ instead of $3$) and the sign in front of $\cos t$.
$u(x)=x+\sin x-xe^x$-integral from $0$ to $x$ of $(e^x(x-t)-\sin x)u(t)dt$