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Find the general solution ofthe given differential equation, 2)2Give the largest interva over which the general solution defined_ (Think bout the implications any s...

Question

Find the general solution ofthe given differential equation, 2)2Give the largest interva over which the general solution defined_ (Think bout the implications any singular points, Enter vout answer using interval notation:)Determine whether there are any transient terms in the genera solvtion_ (Enter the transient terms as comma-separated list; tnere are none enter NONE_ Hint: Expand the genera solution into additive terms; transient term defined as; term as * -> 0 )

Find the general solution ofthe given differential equation, 2)2 Give the largest interva over which the general solution defined_ (Think bout the implications any singular points, Enter vout answer using interval notation:) Determine whether there are any transient terms in the genera solvtion_ (Enter the transient terms as comma-separated list; tnere are none enter NONE_ Hint: Expand the genera solution into additive terms; transient term defined as; term as * -> 0 )



Answers

Find the general solution of the given differential equation. Give the largest interval $I$ over which the general solution is defined. Determine whether there are any transient terms in the general solution. $$\frac{d y}{d x}+2 y=0$$

In this question here we are given X squared Y prime plus Xy is equal to one. And rearranging this into standard form. We have Y prime dividing everything by X squared plus. Why X took both -1 Is equal to X to the 4th -2. And finding our integrating factor Which is E to the power of RPX in this case which is X to the power of the integral of extra. Both -1. D X. We have our E to the integral of one of our X dx which is um eat both lean X and equal to x. So we have our integrating factor as X and much playing everything here. By integrating factor we get X. Y prime plus Y Is equal to X to the 4th -1. And here we can see that the left side is equivalent to the derivative of x Y Which is equal to hear X to the 4th -1 and integrating both sides with respect to X. We get x Y is equal to the integral Of one over XDX. And here we have X. Y is equal to the integral of one of our X is an off X plus constant of integration. See now dividing everything by X. We get our Y and the general solution will be Y is equal to one over x. Lean of eggs Plus CX 2, 4 -1. And this will be our general solution to the equation. Now we have um here we have a term where one of our eggs and we can see that one of our eggs approaches zero is X approaches infinity, And the same with C. X. to the government as one where, yeah, This term approaches zero as x approaches infinity. So this solution is transient and defined from X is equal to zero to infinite. Since we have a kleenex here takes is only positive.

In this question. We are given the differential equation X. Y. Prime plus one plus X. Why? Yeah is equal to E. To the power of minus x. Mhm By sign two x. Now putting this into standard form who have Y. Prime plus one plus X over X. Since we're dividing everything by your ex Is equal to X to the 4th -1. It is both -1 sang two X. No. Here we find our integrating factor is E to the integral of one plus X. Oh my X dx is equal to E. The integral of one of my ex Plus one DX. Here we get E three. No fix a play by Yeah affects. And this gives us X. E to the power of facts. So this is our integrating factor. Multiplying everything here by our integrating factor. We get X. E to the power of X. Mhm. Why prime plus here the excess council we left it one plus X. E. to the 12 x. Why is he go to? And here we have X to the fourth minus one. And ex uh canceling out eat both minus X. And eat of X cancelling out. So we are left the sign two X. Now we observed that the left hand side is equivalent to the derivative of X. E. X. Our integrating factor. And why? Yeah. And here People into the right hand side which is just two. X integrating both sides with respect to X. On the left we have X. E to the X. Y. Is equal to the integral of sine two X by the chain rule the same 1st. Uh the integrity of the sign gives us negative course two X apply by the derivative of the inside divided by the river motive. Here we get 1/2 and we add the constant of integration which is C. And dividing both sides by X. E. To both. X gives us hawaii minus 1/2. Yeah. X. to the 4th -1. E. To the post minus X. Of course mm cost two X. Mhm Plus see X to the both -1 and E. two. The both -X. And this is our general solution. And we can see that since we have divided all sides by X&E We have access to both -1 and E to the -1 on all our terms. This means that this solution is transient because it's X. As X approaches infinity, the value of the terms approach zero. Now this is do you find over the interval of zero to infinity?

Here we are given the differential equation X plus one. Do I TX Plus X-plus 2? Why is equal to two X and E. To the power of minus X. Now, first we re arrange this and put it in standard form we have to I D X. The coefficient of just one. So divide by X plus one X plus one goes here we have experts to Over X-plus one. Played by Y. He had writing bags plus one. We have two X eight of minus X Over X-plus one. Now I want e our about integrating factor will be given by E. To the power the integral of X plus two over X plus one TX. And here to integrate simply first we have to divide X plus two by X plus one Accent to execute a one X-us one. Here we get zero. Here we get one which which means we have one plus one over X plus one And replacing that here we have e integral of one plus one over X plus one TX giving us Yeah. E to the power of X. Apply by E. Lean off X plus one. And this gives us X plus one and E to the poor facts. Is our integrating factor. Now, multiplying everything here by our integrating factor. We get X plus one. E to the power of X. Do Y D x plus here not applying the X plus one council out we'll have to explain to mhm. E to the power fix. Why is you go to hear the X plus one council out? We are left with And the E. To the minus six as well separate each other out. And then we have two X. Mhm Right. No. Here we observe that on the left hand side this is the the derivative is equivalent to the derivative? Uh huh. X plus one eating 12 X. And why? Now the derivative? And this is equivalent to the right hand side Which is three x. And integrating both sides with respect to X. Here we get the left hand side we have X plus one. Each of the four Fix. And why being equal to two X squared over to plus C. And you simplify this? We have these two cancel out. So we have X squared and Over X-plus one E. to the power of -6 plus C. Over So C. E minus to the minus six Over X plus what? And this is our general solution to the given differential equations. Mhm. Now we observed that this solution is transient and it is it is defined when In the range -1 to infinity. So that that line is never getting a negative there. So this is our final solution. Mhm

In this question we are given white prime plus three X squared y is equal to X quipped. Now this is already in the standard form therefore our integrating factor is given by E. To the power of the integral of P X. In this case this is three squared TX ignoring the concept of integration here we get E to the power of X cubed. Now not applying everything by our constant of integration we have E to the X cubed applied by Y prime plus three x squared eat the ball off execute why Is he co two X squared E to the power of X cubed. Now we notice that the left side is equivalent to the derivative he X cubed. Why? And this is Echo two x squared eat of execute. Now integrating both sides with respect to X. Yeah on the left we get e x cubed why is equal to the integral X squared eat both execute dx Now we notice that the derivative of eat both execute is equal to three X squared E to the both execute. Therefore that means that the integral of these Will be in court to hear we divide by three, we divide by three. Therefore here we have 1/3 and this his E. T both execute plus our constant of integration. See therefore dividing both sides by each of both execute here we get Our Y will be called to one of three plus C mm to the post minus X cubed and this is our general our general solution to this differential equation. So I'm going to rewrite it here. Why is he got 1/3? 1/3 plus c. E. O of minus X cubed. And here we can see that we have a transient term here because as X approaches infinity, our C. E to both minus execute minus X cubed approaches zero. Therefore C. E. to the 12 miners execute is transient and our solution is valid or defined over all of X or infinity to infinity. Mhm.


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