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Obtain the general solution to the equation:(3t + Sy + 3Jdt - dy =0The general solution is y(t) =ignoring lost solutions , if any:...

Question

Obtain the general solution to the equation:(3t + Sy + 3Jdt - dy =0The general solution is y(t) =ignoring lost solutions , if any:

Obtain the general solution to the equation: (3t + Sy + 3Jdt - dy =0 The general solution is y(t) = ignoring lost solutions , if any:



Answers

Find the general solution of the differential equation. $$y^{\prime \prime}+y^{\prime}+3 y=0$$

Okay, so for this problem let's go ahead and start off by replacing these white terms with our terms instead. So have R squared plus three. R plus two equals zero. From there. We can go ahead and do a little of factoring and we should get our plus two Times are Plus one and this equals zero. And then from this we can extract that are our values are negative one and negative two. With these are values. We can go ahead and build our solution solutions of the form Y equals to see one, eat the negative X Plus C. to eat the -2 x. And so this is our answer to the differential equation above.

We have the second order differential equation, Y double prime plus two. Y prime plus three, Y equals zero. So we have coefficient in front of the double prime term is already one. So you don't need to divide through by it. So we have bs two CS CS three. So we have b squared minus four. C. Is four minus 12 which is minus eight which is less than zero. So it's best to write it in, trigger the metric for him. Um Now our character roots of our characters, the equation are going to be complex numbers so this is minus B plus or minus square root of eight or squared of minus eight. Um So which is then to square two times I all over too. So you get minus one plus or minus square to I. So to construct our solution then we take this that's either the minus. So we get either the minus T. And then this is our coefficient of our signs and our coastline terms. So a co sign square to to T. Plus B. Sine squared of two T. So you can see are two solutions are you know this square to to T. And then either the minus T. Co. Sine squared of two T. Those are the two general solutions. And then we take a linear combination and then I factored out the E. To the minus team. Yeah so again it's writing it in exponential form. You can do that unless he here either the listener, she's just right here. Either the -1 plus square to to IT. And I don't want to use the same constants because they won't do the same values. Um into the -1- where the two. I can't. That is mathematically equivalent to this but. And what we'll find is that you can notice these. If you kept it this way you're gonna say well, wait a minute, shouldn't shouldn't the result of this be a real number? Well, it will be because if you have initial conditions notice these are complex con jackets or one another. And so what will happen is these will be complex con jackets at one another. So you take this whole term as a complex conjugated that, and when you add complex condo gets together, the imaginary part, it drops out. So in the end, if you would have initial conditions and applied them, you'd see the C and D. Are going to be complex consequence of one another. So in the end you'll you'll get real numbers out of this, which is not obvious to begin with. And that's why a lot of time basically write it as this form. So we don't have to deal with the fact that we're we've already used the fact that we're going to get complex consequence

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.

Here we have a second order differential equation and they want us to find a general solution for that. So we can identify B. Its coefficient of the first river term and that's minus three. And C. Is the coefficient of the Or the zero derivative term or just the wide term. And that's too. Then we can calculate B squared minus four C. And that turns out to be one which is greater than zero. So writing this as an exponential, an exponential form will be the most convenient for him to write it. All the forms that they give in the book are actually mathematically equivalent as long as you as long as you are happy to deal with complex numbers. Um They're they're all mathematically equivalent but you have to you just need complex numbers to make them. So so we can just say, you know, use the characteristic equation is one is one. You know, is r squared minus three, R plus two equals zero. Right? And so we can find the roots of that. So we get minus B. Three plus or minus this thing here, B squared minus four or a is one, so B squared minus four C. So this is 1/2. And so we get R one and R two. There are two routes are one or two and one. So we know we have either the tea and eat of the two T. S. Solutions. And so any linear combination of those will also be a solution. So we have our general solution as a way to the T Plus B, either the two T. And um we can find A and B if we're given some initial conditions or boundary conditions.


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