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Given the following differential equation.y''' + 9y'= e4t + sint1) Find the general solution of the corresponding homogeneousequation.2) Find a ...

Question

Given the following differential equation.y''' + 9y'= e4t + sint1) Find the general solution of the corresponding homogeneousequation.2) Find a particular solution of the nonhomogeneous equation.Briefly define and describe the method that you use.3) Solve the nonhomogeneous equation.

Given the following differential equation. y''' + 9y'= e4t + sint 1) Find the general solution of the corresponding homogeneous equation. 2) Find a particular solution of the nonhomogeneous equation. Briefly define and describe the method that you use. 3) Solve the nonhomogeneous equation.



Answers

Write the given nonhomogeneous differential equation as an operator equation, and give the associated homogeneous differential equation. $$y^{\prime \prime \prime}+x^{2} y^{\prime \prime}-(\sin x) y^{\prime}+e^{x} y=x^{3}$$

Okay. So for this problem we're gonna we're gonna have to find the original differential equation. And so to start off, let's go ahead and identify our values sir. Our values in this case since it's co sign and sign, We're going to be negative three I. Sorry plus or -3 i. And from that we can go ahead and drive that are sorry there are factors. Original factors would have been R plus three I Times are -3 i. And it would have equalled zero. So I have R squared minus three. I plus three I R 33 I r minus nine I squared And it would equal zero. So we're to have R squared Plus nine equals to zero. And if we go ahead and replace this R squared with white terms instead, well have Y double prime plus nine Y equals zero. So for this question this is going to be original differential equation and our answer.

Okay lets surface problem by replacing the Y. Prime bye. D. Y squared. Or sorry D. Y. Divided by T. T. And that equals the negative sign T. And now we can move the D. T. Turn to the other side so of the three Y squared dy equals two negative science T. T. T. Now we can integrate both sides. So on this side we'll have that Y cubed equals to co sign T. Plus C. Or rather than why equals to the cube root. Of course I 90 plus C. And now we can use the initial conditions that we had which were why of poverty vehicles one and so where we see a wide term we're gonna plug in one and where we see a co sign or rather start a. T. Term. We're going to plug in a pi over two. So we're gonna have co sign of pi over two plus C. And co sign of pi over two is zero. This entire term is gonna go to zero. So we're going to have one equals to the cube root of sea. Or rather that C equals to one or one cubed. And so now we can actually build our final solution. So our final solution is going to be Y equals to the cube root. Co sign T plus one.

Okay for this problem we are asked to find the original differential equation. So let's go ahead and start off by finding the R values. They are values in this case are going to be the form A plus B. I. Since we have co science and science but and the A for both of these are going to be zero. So we're just going to focus on the I. B. Term. So if the I in the first case is going to be plus or -1 are sorry did the B in the first case is gonna be one. So we're just gonna put your mind's eye And the second case it's going to be too so we'll have plus or -2. I so these are our values here and from that we know that we have the roots which is gonna be our plus I ar minus I R. Plus two I And AR -2. I It's all equals to zero. So let's go ahead and expand this out. I'm going to focus on the first two right here first, so r squared minus ir plus Ir plus or minus I squared. That's gonna be left to taken care of On the right to We'll have our squared -2 ir plus two Ir RC -4. i squared. All right. And we can go ahead and some fights out just a little bit more. We know that these two terms cancel over here. So we're left with R squared plus one. And then the right hand side were left at the r squared plus four. So we can expand this out just a little bit more. We'll have our to the 4th Plus four, R Squared plus R squared plus four. And so we have art of the 4th Plus five, R Squared Plus four. All equals to zero. Okay, and from here we can go ahead and replace our our values with Y values. So for here we're going to have why The 4th Row to Why? Here we'll have Y double prime and here we'll have why? Mhm. And so this is going to be original differential equation and our answer.

Okay so surface problem by replacing these Y terms with our terms instead. So we'll have to R squared plus three. R plus one equals to zero. And from here we're gonna do a little bit of factoring so I'm gonna go ahead and rewrite this as two R squared plus two are Plus are plus one course zero. The left hand side. Or rather this right here You can factor out of two are so you have to our times are plus one and then the right hand side which would be this right here We can just factor out of one. I will have our plus one and so we can have two different factors who have to R plus one times are plus one and this equals to zero. And from this we can extract that are our values are negative one half and negative one. Yeah. And so with these are values we can go ahead and um build our solution. So our solutions of the form Y equals two. See one Eats the negative T plus C. To eat the negative. She divided by two. So that's your answer.


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