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[40 marks]Solve the following second order nonhomogeneous differential equation: +2y 0.75y cosri 025sml >) 4 0Wr J(0) 2.78 V(0) =...

Question

[40 marks]Solve the following second order nonhomogeneous differential equation: +2y 0.75y cosri 025sml >) 4 0Wr J(0) 2.78 V(0) =

[40 marks] Solve the following second order nonhomogeneous differential equation: +2y 0.75y cosri 025sml >) 4 0Wr J(0) 2.78 V(0) =



Answers

Solve the following differential equations: $$y^{\prime}=e^{4 y} t^{3}-e^{4 y}$$

Hello Friends. We have to solve this given differential equation. Due to the power for by -5. The script let's for weight goes to zero. So we can at the House of Education. And to depart for minus five. M squared plus four N. equals to zero. Yeah. Yeah. So we cannot escape plus four. Okay. And mm square. Sorry This will be a musical municipal multiplied way Mm Scar -1. So money's to do is manage to Entries minus entry is one. And um food is -1. So why will be close to 7? It is the power of m. n. x. Plus you to E. To the power of M. Two X. Let's see three TV power f. m. three x. Let's see four into the power of and poor X. So why will we close to 7? Stephen into the power to X. Plus You to Eat? To the Power of -2 X. Plus the three into the power X. Let's see four into the par -X. So this is that solution of this given differential equation. I hope you understood

At all. We have to solve this given differential equation That is D. three by upon. Mhm. The excuse minus two need to buy upon D. X squared minus of Mhm. Three D. Viable. Do you expect us to zero? So we can write it? D. Q minus. Do do you squared minus of three D. Very close to zero. So auxiliary question can witness MQ -2. Emmy Square -3. a.m. Cause two Zeros and will be common. So mm squared minus two M. Monastery close to zero. So we can write em into I am Monastery and plus one Because two Zeros MLB Mm zero M. 2 did he? And M. Three. Yeah mine is fun. So solution came readiness by Costas even into the power of And my next policy to into the power of M. two x. Let's see three M. 3 U. to the power and three x. So this will be cause to fight cost to siva into the power of zero and two X plus E. To into the power three X Plus C. three into the power of -X. So why will be caused to seven Bliss C. two into the power of three x. Let's see. Three into the power of -6. So this is the Answered. I hope you understood.

So start this problem trying to solve for the homogeneous equation. So we'll have that R squared plus are equals to zero or that we can fact this out to be, our times are plus one equals zero. With that we get our values of negative one and zero. And so now we can actually build our homogeneous solution. So homogeneous solution is going to see one, E. To the negative X plus C. Two. And we can also take a stab at our particular solution guests. So our guests for the particular solution is going to be to form a X plus B. But notice that there's these two terms that are similar power in terms of the variable that they have and we don't want that. So what we end up doing is we end up multiplying by a factor of X in the guests for a particular solution. And so what the ends up giving us is we have a X squared plus bx as our guests for a particular solution. And now we take the driver this twice. So I have to A X. Cosby and the second river is going to be to A. And then we can plug these equate are these equations into our initial equation that we had which was Y double prime plus why prime equals to four X. So in this case, wide old primaries to A plus Y. Prime, which is to a X. B plus B. This equals to four X. And so now we can create a system of equations. So let's do the like terms first, which are these two right here and these two So we have A to A plus B equals zero and that to A equals to four. So with this we can get that A equals to two. We can plug this into our initial equations. We have the two times two plus p course zero For the B plus four equals zero or that B equals two negative four. And so with this, you can actually build our total solution, which is the sum of her homogeneous solution plus the particular solution. And so our total solution is gonna being C one E. To the negative X plus C. Two uh plus two X squared minus four X.

All right. So this time we are working with the differential equation of the second, distributive of why -4 times the first derivative plus four times Y. B zero. And so this one is a little bit unique because when we factor it for working with the auxiliary equation, which is going to be r squared minus four R plus four. Factoring this gives us ar minus two square, right? Because um negative two times negative two of those A positive four. And then anyway, so we have is that our our one is equal to R. R two which is equal to two. So when we go to put it in our general solution, we're gonna do something a little bit different uh in the past few problems. So we'll take Y and we're going to take the first step is the same and we have a constant times E. To the power of their first are two times X. And we add that to a constant. But this time due to the multiplicity, we're going to take that constant times X. And multiply that by E to the power of two X. Okay, so the general solution has this extra X in here. So it's good to be aware of this property when solving these differential equations. Otherwise, this is going to be our journal solution, since we have no initial values to solve it any further.


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