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Determine the longest interval which the given initial value problem certain have unique twice-difterentiable solution. notation _ 9)y" - 7ty' + Sy sin L,...

Question

Determine the longest interval which the given initial value problem certain have unique twice-difterentiable solution. notation _ 9)y" - 7ty' + Sy sin L, Y(-4) =

Determine the longest interval which the given initial value problem certain have unique twice-difterentiable solution. notation _ 9)y" - 7ty' + Sy sin L, Y(-4) =



Answers

Solve the given initial-value problem. Give the largest interval $I$ over which the solution is defined. $$y^{\prime}-(\sin x) y=2 \sin x, \quad y(\pi / 2)=1$$

In this question we are given that differential equation. L. D A D T plus R. I. Did you go to E. And the initial point I was going to die. Not when T is equal to zero. Now putting this into rearranging into standard form, we divide by L. Everything by L. You get the I D T plus R L. I. Is he go to E over here? No, we can now find the integrating factor which is given by E. To the power of the integral of ara of R L T T. Which gives us E to the power of Paraty over here. Now, I'm not playing everything by the integrating factor. We get E to the power. Arab T R. L apply by the I T T plus R A E. To the power of Paraty over L over L apply by I is equal to mm And eat of Paraty over L divided by L. No, we observe that the left hand side is equivalent to the derivative of E. To the power of charity over L. Multiply by I is equal to the right hand side which is E and E. To the ball. Paraty over L divided by health. Now integrating both sides with respect to T. Here we get E to the power of Paraty over L. Applied by is equal to here. We have our E over Ls Constance so E to the power of Arab T over L. The integral of this gives us So our concerns already we have E over L And the integral E. To the polarity over L gives us L over Sarah more played by E. To the power of arrow. T over L plus the constant of integration. It will be seen. Mhm Now here we have an E. And into the power parity over L. Here dividing everything by this we have I is equal to here. We notice that this cancel out. So we have E over our and this cancels out with this on division plus see and E to the power of arrow. T over L. But this is dividing so we get minus over there. So this is the general solution for the given differential equations. Now to find the specific um solution. Now we are substituting our initial values right here so we have I being equal to high note yeah is equal to E Over our plus see and here our T will be going to zero. So we have eaten pope zero. Now this will give us our C. Is I not minus E over A Mhm. And this is our constant of integration. The value of the constant of integration and substituting this into our main equation. We are going to have our final solution is our final solution is going to be high is equal to mhm Over her Plus our see that recalculated. High not minus E over our want to play by E. To the poor minus R. A T. Over. Yeah. And this is our final solution which is defined from minus infinity to infinity. Mhm mm hmm.

Hello and Welcome to 15 of Chapter two. Section 4 you were asked to solve the give initial value problem and determine how the interval in which the solution exists depends on the initial value of why not? So are Given initially problem here is why Prime Plus White Cube zero. We can simply move the white cooped over and divide it. So we'll have could do I. T. Two is equal to negative Y cubed. From here we can bring the weight down. DT up and simply integrate will be left with. Dy over negative. Why cubed is equal to D. T. I'm gonna bring this negative to the other side and drop that here. We can just integrate so integrated facades. Great. So we'll have um the integral of dy times Why it's the -3. Well what is the integral of wives and negative three D. Y. Well that's um why to the negative too over negative two. And that's going to be equal to uh the integral of negative DT. Which is negative T. Plus C. Okay and from here we can just solve for Y. And we'll do that by multiplying everything by -2. So let's move this up here. We'll have why it's the negative too is equal to Well to C. Plus C. And remember this is actually one of her Y squared one over Y squared. So we're going to bring this to to proceed turned down the Y squared term up and then take the square root. Lastly we'll be left with Y. Is equal to Uh plus or minus the square root of one over to t plus six. And that is our um initial does the solution to our initial value problem? However, we do have this um initial value so plug that in. Um it's a race down here a little bit. So if we plug in, our initial value will be left with drink, why? Why not is equal to plus or minus The square root of one, oversee. Where we can take the square. We can square both sides will get why not squared is equal to one oversee. Yeah, such that C. Is equal to one over. Why not squared? Great. And from here we can yeah from here we can substitute uh this one over why not square term for safe back in to our solution and it's right here. Let's see. And then we'll have one over. Why not squared? Great. And looking at this, there are uh we have to find out where uh Sorry what? Why not? Cannot be. And um just by simply looking at it, we know that we cannot divide by zero. So why not? Cannot be zero? Let's uh let's write that down here. Why not cannot be equal to zero? Mm hmm Let's also just rewrite the um solution so we'll have y is equal to plus or minus square root one over to t plus one over. Why not squared? Okay, so we know why not cannot be equal to zero. We also know that we cannot divide by zero in this case. So 22 Plus one over. Why not squared cannot be equal to zero. And um lastly yeah we cannot have a negative sign underneath the uh square root. So we'll have Um one over to T. Plus one over. Why not squared must be greater than or equal to zero. So the first one is already solved the second one we can um consult this for why not? So subtract two T. Some black subtract two T. And one of her. Why not squared People to -2 T. Um We can then just solve for or why not? So get why not is equal to the square root of Well actually cannot be equal to of um 1/-2 T. So that's another constraint. And our last constraint is this is going to be a little bit more challenging to figure out. But yeah 1/22 teeth whatever by square must be greater than equal to zero, which means this denominator must be positive. So we'll write this as to really just the same things up there. But to to plus one over. Why not squared must be greater than um zero. And we really solve this. We just solved it up here so we can write this as why not? Let's be greater than the square of one over negative two tape which essentially encompasses that one. There are two real constraints for why not or that cannot be equal to zero. They cannot pick 20, and it must be greater than the square of one over negative two tape, and that concludes our problem.

Hello there. For the following exercise we have the following initial value problem. So the day is the first solve this um differential equation and then evaluate at the initial value to obtain the value of the constants because first we obtain a family one parameter family of solutions for this differential equation. Okay so let's focus on that equation. And it's clear but these are separate differential equations. This is equal to sin X. The x minus Y. Do what? Then we integrate to both sides. And we obtained here minus cosine of X equals two minus y squared over two from this. We have that why? The absolute value of why is equals To the square root of two times co sign of X and plus some constant here plus C. Okay but we have an initial value problems so we need to Evolution or function at zero and why will be equal to one. Okay. And we're going to do this in order to obtain the value of the constant here. So that implies That Y will because to one And the value of the integral will be a square root of two. And of course he is here the plastic And from this Well it is clear that c It's going to be equal to 1- the square root of two. One that we have to find the constant. We need to replace the final solution. And we obtained that the absolute value of Y is equal to the square root of two times cosine x Plus 1- the square root of two


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