To start solving this differential equation. I'm going to rewrite this differential equation in terms of its linear operator form. So it's gonna be D squared plus nine. Why isn't be able to five co sign two X Now we can get the corresponding complementary solution by setting the right hand side with zero so b squared. Why carpentry is he with zero? So now, to solve this, we find that the auxiliary equation p of our isn't B r squared plus nine is equal to zero. So we get that are is going to be able to plus or minus three I This gives us a complimentary solution. Why of X equal to C one e to the, um sorry. See one sign of three x plus C to co sign of three X Now, to get rid of this right hand side here, we're going to find its annihilator. So after Lex is equal to five co Zain two X So then, uh, the Annihilator. Okay, we have here that b is equal to two. Also, A is equal to zero. So remember, our Annihilator is gonna take the form The squared minus two a D plus a squared plus B squared. So we get that, um so 80 So this just becomes here. Then we have two squared, which is gonna be equal to four. So we get these squared plus four as our Annihilator. Now we apply our Annihilator toe our original equation here. So we get t squared plus four and then B squared plus nine of why this is equal to and then a pie are Annihilator to this side here and that will be equal to zero. K is equal to zero. Now, solving this equation here our auxiliary equation p of r zero to R squared Plus for our square plus nine is he with zero? So we get our sea with you plus or minus two I and then our is equal to plus or minus three I. So the plus or minus three I were to get from our complimentary solution So you have and why of X would be equal to see one sign of three x plus C to co sign of three X and then we're gonna have a plus now the plus or minus two I that's going to give us on a not sign of two X and then plus B not Oops. Be not cosigned of two X. So next we need to solve for a not And why not? This is our trial solution here. So we're gonna plug this into our original equation differential equation here and then find then use this to calculate a not and be not so. First, we need to find the second derivative. So Y p prime first. Why? P prime of X. This is gonna be able to to a not co sign of two X and then plus to be not Ah, sorry. Minus to be not sign of two X Now, why P double prime of X is going to be equal to all right is equal to, um, sir, Right by P double prime. So we get negative, right? For negative for a not sign two X, uh, sorry, minus two. Ah, sorry. Four b not cosigned two X like So now we can plug these into this formula. Here. It's We get negative for a not sign. Two x minus four B, not cosigned two x and then plus nine times a not or nine times this. So nine a not sign two x then plus nine b not cosigned two x and remember, this is equal to five co sign of two X So combining our sign terms Here we get, uh five A. Not to get five, a not sign of two X and then plus five b not cosigned of two X is equal to five co sign of two X notice here that there's no signed her plus zero sign. So that means five a not is gonna be zero. So we had five a Not is he of the zero And then we also get that five b not is going to be able to five So you're not zero and be not is equal to one. So since a not a zero Weiqing uh, cancel that out and be not distressing with the one. So this is going to be our general solution here. Now we need to use our initial conditions to get, um, this C one and C two. Okay, so that's right. Our initial conditions again. So why double prime plus why a nine y his people to five cosigning two x okay. And then our initial conditions Why zero is equal to two. And then why Prime a zero is equal to three. Okay, so our general solution this why of X is equal to see one. See one signed three x plus C two co sign three x and then plus cosigned two x. So let's plug in. Um, Well, first we need to find why Prime. Oh, are. Actually, we can go ahead and plug in zero here, we'll get why of zero is equal to so sign of zero is zero. So we'll end up getting C two and then, um co sign of zero is one and then co sign of zero is also one. So get C two plus one is equal to two. So that means C two is equal to one. Okay, so now that we know that C two is equal to one, we can, uh, plug it back in here. Okay, so that's just gonna be one here. Now we need to find why prime? So we can use this. So why prime of X is going to be equal to So now we're going to use, um Okay, that's gonna be three c one sign of three X. Well, just like that are sorry. Co sign Coastline three X and then we're gonna have minus three. Sign three x minus to sign two X So then plugging in. Why? Prime of zero or at zero? So co sign of zero is gonna be sorry. Coastline of Zero is gonna be one. This is zero and zero. So we just get three C one and that's equal to our initial condition here, which is three. So we get that C one is also equal to one. So our final solution, it's just going to be this. Ah, this equation here.