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2 . (20 points) Use variation of parameters to find the general solution to the differential equation ty" + (1 _ 2t)y + (t - I)y = tet given that the functions...

Question

2 . (20 points) Use variation of parameters to find the general solution to the differential equation ty" + (1 _ 2t)y + (t - I)y = tet given that the functions Y1 et and 92 et Int are linearly independent solutions to the corresponding homogeneous equation for t > 0_

2 . (20 points) Use variation of parameters to find the general solution to the differential equation ty" + (1 _ 2t)y + (t - I)y = tet given that the functions Y1 et and 92 et Int are linearly independent solutions to the corresponding homogeneous equation for t > 0_



Answers

Find both the general solution of the differential equation and the solution with the given initial condition.
$$
\begin{array}{l}{\mathbf{r}^{\prime \prime}(t)=\left\langle e^{2 t-2}, t^{2}-1,1\right\rangle, \quad \mathbf{r}(1)=\langle 0,0,1\rangle} \\ {\mathbf{r}^{\prime}(1)=\langle 2,0,0\rangle}\end{array}
$$

Okay, so we can get the generously shin bi integrations. That's the interval of E to the power T d T car the integral off the sign of t g t. So the 1st 1 is e to the power of tea plus C one common negative co sign of t plus two C two. So we know that at zero we have zero comma to zero results in zero comma to. So if we plug it in, we end up with C one is equal to native one and see to his equal to three. So this is your general solution than if you just plug in numbers. You end up with your final solution off t to the power of TV minus T Come on. Three T minus the sine of T minus three.

Right. So let's sort this problem off by solving for that module this solution. So I have R squared plus ar minus two. R equals to zero. And it looks like we can get our plus two times arm and so on Equals zero with some factory. And so this gives the solutions of our equaling one and -2. And with that we can build our homogeneous solution. So homogeneous solution is going to see one E. To the X plus C. Two E to the negative two X. And we can take a guess at our particular solution. So our guests for a particular solution is going to be X plus B plus see co sign X plus D. Sign Sorry, Co signed two x and decide to X. Right. And we have to take the driver of this particular solution guess twice. So the first drift is going to be a -2 C signed two x plus to the Co signed two x. And the second derivative is going to be negative four C. Co signed two x -4 d. sign two x. And right now we have to plug this into our original equation. Our original equation was wide of a crime plus Y. Prime minus two Y equals two X plus sign to X. And so we substitute into the original equation. So we'll have negative four C. Co sign two x -4 d. sign two X plus Y. Prime which is a minus two C. Signed two X Plus two d. co sign two X minus two times Y. So we're going to have to A X plus to be we're sorry this is a minus. So be a minus here um -2 C. co sign two x -2. The sign two X. And this all equals two X. Plus sign two X. And so let's try and sympathize out so we can have a cleaner system of equations. So I see life terms here and they write this in a different color. Actually you like terms here here here and here. Yeah. So We will end up with negative 60 Co signed two x minus 60. Sign two X plus a -2 C. signed two x plus two D. Co signed two x -2 X -2 B. Equals two X. Plus side two X. And so with that let's build a system of equations here. So let's see if we can get some like terms let's have the co signs first. So we're gonna have negative six C. This term this term contains of co sign so negative 60 plus duty equals to zero. And let's do signs a negative 60 minus two C equals to one. And then let's do this term right here. So we'll have negative to a. He goes to one and finally we'll have a minus to B equals to zero. Right so right off the bat we can solve for a here. So we'll have a equals 2 -1 half. And with that we can solve for the bottom equation. So we'll have a negative one half minus to B equals to zero. Bring the one after the other sides will have negative to be equals to one F. Or b equal to negative 1/4. And now let's try and solve for the top two equations. So let's focus on those um negative 60 plus two D. Equals to zero And -2 C -60. He goes to one. So let me multiply the Top equation by three. We'll have negative 60 plus two D. Equals to zero negative six C minus 18 D. Equals 23 Let's subtract these terms. We can eliminate the seas And we'll have 20 d. Equals 2 -3 or D equals the negative 3/20. And with that we can solve for C. So we have that negative 60 Plus two times negative 3 20 equals to zero. And so then we have negative 60 plus. Actually it's gonna minus six. Or other sorry 3/10 equals to zero. And let's bring the 3/10. The other sides will have negative 60 Equals at 3/10. And if we divide by negative six we get C equals two -1/20. Right? And now we can build um are total solution. So our total solution consists of our particular solution closer homogeneous solution. So we're going to have c. one E. To the X plus C. Two E. To the negative two. X minus one. Fourth minus one half x minus 3 20. It's signed two x minus 1/20. Co sign two X. And with that we can solve for C. One and C. Two using our initial conditions. So our initial conditions are Why have zero equals 1 And why prime of zero go 0. So let's start off by taking the direct route of this equation at the top. So we have that Y. T. Prime equals two. C. One E. To the x minus two C. To eat the negative two X -1 4th. or rather sorry this is a constant. So it cancels out. So I have minus 1/2 minus 3/10. Sign sorry Co sign two X plus 1/10. Sign two X. And so our initial condition for the first equation is Y zero. So it's plug in Y. Of zero in the first equation. So we'll have that see one plus C two minus 1/4. And when you plug in zero for sign it gives you zero. So we'll have that turns to zero and we'll have minus 1/20 That are equals to one. And her second equation is going to be when we plug in zero. So I have C1 -2. C two minus one half minus 3/10. All equals to zero. Right so now we can solve for C1C2 with the system equations. So let's start off by simplifying these constant terms. So we'll have that C one plus C two equals two, 3/10. And for the bottom equation we have C one minus two. C two equals two. 8/10. Uh huh. Right so now Let's subtract this so we can cancel at c. one. So I have 3 C2 Equals, two, or 10. Where that C2 Equals so negative 5/30. Sorry about that. I think I made a mistake here actually in the math this should not equal 2 3/10. In fact It is 1 -3/10. So it's simply going to be 7/10. So with that everything else should work out. So have This being 7/10 And the bottom occasional some fly down to see one -2. C two equals two. Excuse me, I made a mistake again. It's not 7/10 is 3/10 plus 20th or one equals 13 10th. All right. And the bottom occasional simply down to 8/10. So with that let's do some subtraction to cancel. Let's see one. So we'll have that three C two equals two five over 10 and then C2 equals two 5/30 Which sympathize down to one six. So with that we can solve for C. One. So let's use that C one plus C two equals 2 3/10. So I have that C. One plus C two Equals 2/13/10. And let's move the 16 The other sides will have the 13 10th minus 16 equals to see one. And so C1 will equal to 39/30 minus 5/30 which is 34/30 which flies down 17/15. Right? So now we can actually build our total solution which let's remind ourselves is the sum of the homogeneous solution plus the particular solution. So our total solution here it's gonna be 1/6. He to the -2 x plus 17 15. E. To the X plus or rather minus 1/4 minus one half. X minus 3/20. Signed two X minus 1/20. Co sign two X. Right? That's it.

To start off this problem, we will want to divide off the two in front of the wide of abrupt. This will leave us with my double. Prime is equal to zero Now we will want to integrate both sides. These are indefinite integral. So be careful of the constant. The integral of why double prime will be wide prime plus a constant and the integral of zero will be zero plus a separate constant. These constants ca NBI added together and we can just have a new constant which we can label C two on the left side. Now we will want to integrate both sides again. When doing this, the integral of UAE prime will be why and the integral of C two will be C two. I'm sex now Will you will need to add on another constant. The integral of zero is zero plus a separate constant. Again these constants can again be combined into a different constant weaken label See one Now we will want to get why alone To do this, we will want to subtract off see two times X and C one This will leave us with y equals negative C two times X minus C. One. Because thes constants are unknown, the negatives can year raced. This is because when doing the algebra behind solving for them, the negatives would cancel out if it was a positive, and if the constant needed to be a negative, the constants would show negative, no matter what. Writing it, as if the constant were positive makes it easier to see what the equation is.


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