To solve for the original equation, we're going to take the integral four times, so the integral of Y to the four TT equaling the integral of the quantity -4 sine t + 2 cosine TDT. We're going to get Y to the third prime equal 4 cosine t + 2 sine t + C one. Sticking in the original .0 for T and 10 for Y triple prime, we get 10 equal 4 cosine 0 + 2 sine 0 plus C1. Cosine of 0 is 1, sine of 0 is 0, so we get C1 being 6. Now we're going to take the integral of the Y triple prime DT equaling the integral of four cosine t + 2 sine t + 6 DT. So Y double prime equals 4 sine t - 2 cosine t + 6 T plus C2. Now going up and looking at what our Y double prime was, we were know T is 0 when Y double prime is -2 so -2 equal 4 sine 0 - 2 cosine 0 + 6 * 0 plus CT suit C sub two. So C sub two is going to be zero. We're going to take and do the integral again. We're going to get Y prime equaling -4 cosine T -2 sine T plus three, t ^2 + C sub three. Going back to the original, we know that T is 0 when Y prime is -4. So solving we get C sub three equaling 0. Doing the integral one more time, we get Y equal -4 sine t + 2, cosine t + t ^3 + C four. Going back to the original, we know that when T is 0, our original Y is -1. Plugging that in, we get our C4 being -3 for a final answer of Y equal -4 sine t + 2, cosine t + t ^3 -, 3.