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Use the Laplace transform to solve the following non-partial differential equation homogeneous: 02 u fo; dt2 d12where and f are constant; 0 < x < 0 t> 0,an...

Question

Use the Laplace transform to solve the following non-partial differential equation homogeneous: 02 u fo; dt2 d12where and f are constant; 0 < x < 0 t> 0,and satisfying the conditionsdu du u(z,0) = 0, = 0, u(0,+) = 0, lim dl 4_OC d1 t=0

Use the Laplace transform to solve the following non-partial differential equation homogeneous: 02 u fo; dt2 d12 where and f are constant; 0 < x < 0 t> 0,and satisfying the conditions du du u(z,0) = 0, = 0, u(0,+) = 0, lim dl 4_OC d1 t=0



Answers

Use the Laplace transform to solve the given system of differential equations subject to the given initial conditions. $$\begin{aligned} &\frac{d x_{1}}{d t}=2 x_{2}, \quad \frac{d x_{2}}{d t}=-2 x_{1}\\ &x_{1}(0)=0, \quad x_{2}(0)=1 \end{aligned}$$

The first thing we're gonna do here is take our second initial condition and plug it into the first equation that we have. So x one crime at zero is equal to zero is equal to x one and zero plus two x two of zero. You know, the X one of zero is one. Therefore, X 20 is negative 1/2. Now, we're gonna use this later on, but we're gonna take the flash transform of both of our equations to get that this is ask times loss, transform of X one minus x one and zero minus look. Loss. Transform of X one minus two times a plaster transfer next to is equal to zero as well as X or ask times of assurance for the next two minus x two at zero minus two slash France were of X one minus. Cautions for X two is equal to zero and we're gonna group terms together through by factors of x one and X two So that this becomes s minus two times X one three x minus rs minus one times x one minus two x two is equal to x one and zero, which is one, and we're going to ask and negative too. X one plus asked. Minus one times X two is equal to x two of zero, which is negative 1/2. And now we have a system of equations that we can write as a Times X is equal to okay, where X and B are vectors and A's and matrix. So our A is going to be asked minus one negative to negative two and s minus one. Our X is the postulates were x one and with postulates were of X two and R B is just one negative 1/2 right set here and now we can use Cramer's rule to figure out what x one and x two are. So we know that X one will be the determinant. Ah, the one over the determinate of Hey, where the determinant, Uh, a will be at the determinant of s minus one negative to negative two and s minus one, which will be as minus one squared plus four which we can expand to be as squared minus two s plus one minus 40 Gave minus for not bust since negative two times negative. Two is positive. We have to subtract the off diagonal multiplication, which is s squared minus two s minus three, which is s minus three and as post one. So there we have determined into a Now we're going to calculate the determinants of the one which is equal to replacing the first column in our egg with B. So this will be one negative. 1/2 negative too. AZ minus one. When we calculate this determinant, it will be asked minus one minus negative times. 1/2 is just one. So just be asked. Minus two. So therefore X one is s minus two for a clash. Transform with bex. One is s minus you. Over s minus. Three times I supposed one and we're gonna write this as two separate fractions, one for s, minus three and one for X plus one. And to do that, we're gonna multiply through by the denominator. To get that s minus two has to be equal to you. A times expose one plus being times s minus three and foiling of terms. We'll get that a plus being all times s plus a minus three B has to be able to ask minus two. So Therefore, a plus B has to be equal to one or a is equal to one minus B. In the second equation, we have a minus three b. You make substitution that is one minus B. So one minus B minus three B is a negative four. Being is equal to negative two. So therefore, the is 3/4. If a baby is 3/4 than a is one minus Greek borders, just negative. 1/4 no positive, 1/4. So 1/4. So therefore, our the last chance for X one is there for 1/4 times one of the s minus three plus three sports times will never ask. Plus one. Now we can take the inverse LaPlace transform to get that X one of t is equal to 1/4 times Inbursa posture in support of one ever asked minus three plus 3/4. And is the inverse a posh transformer whatever s plus one, then we can use the rule that the class trips will. Each of the A s is he will the one over s minus A. You can notice that in these two inverse applause transforms. Our first are a is three. This will be 1/4 times each of the three team. This should be t plus 3/4 times E to the negative. T That's our X one. You know, we're gonna do all that again for X two. So we know that X two will be the determinant of me too. Over the determinate from a we're gonna calculate the determinants of B two to B the determinant uh, s minus one, a negative to a one in a negative 1/2. So replacing the second column in a with R B, which would be negative 1/2 s just 1/2 plus one, which would be equal to negative 1/2 s plus rehabs or shouldn't be plus one. This is plus her minus native two times. Once a positive too. We should make this fraction five halves and now we're gonna substitute this ends. This is negative 1/2 s plus five halves, all divided by the determinant of A, which was s minus three times as plus one. We're gonna write this as a fraction press minus three in a fraction for us post one. And we're gonna most by through by the denominator to get that negative. 1/2 s plus five halves is equal to you. We had to did this algebra a plus B times s plus a minus three B. It's therefore a plus. B is equal to negative 1/2. So a is equal to negative 1/2 minus being living like it's second equation A minus three be bugging in a will be negative 1/2 minus four. Being is equal to five halves. The ad won't happen. Both sides to get negative for B is equal to three. Therefore be is negative recorders. If B is negative 3/4 this is negative 1/2 which is 2/4 plus 3/4 which is 1/4. So therefore our X two wolf loss transform is 1/4 times one of the s minus three minus 3/4 Times won't ever s plus one. And now we can take the inverse of loss transform to get that X two t is 1/4 tones. Inverse applause. Transform of one of the S minus three minus 3/4 times inverse applause transform of one of her, as opposed one which we already figured out what these two are. So therefore, x two of tea is one border times E to the three T minus 3/4. You need to the negative

The first step here is going to be taking the applause transform of our system of equations. And when we do that, we're going to get s times the thoughts transformed X one minus X one at zero. Well, plus two times of Austrians with next to is equal to zero as our first equation and ask times of loss transform next to minus X two and zero minus two terms of Russians. When the next one minus four terms with restaurants, one next to is equal to zero. Now we're gonna group terms, but x one and X two and pull the constants. Right. So we're gonna have s times across transform X one plus two terms of class transfer. Next to is a well, the X one and zero, which was one. And we're gonna have a negative too. Turns explain. Plus s minus four of the class transform. X two is equal to X to it zero, which is just one. So now we can form a matrix equation with this. So we'll have s to native to you and asked minus four ball times for X one and x two Physical 211 You're going with this being his ex in this A. Therefore, using Cramer's rule, we can figure out what x one and x two are so actual in the Austrians. When Rex one will be the determinant of the one over the determinant of a and we can calculate the determinants one which will just be replacing the first column in A with being. So we'll have 11 to s minus four, which would be asked minus for minus two. Just ask minus six. We're gonna calculate the determinants of a just determining s too negative too. And s minus four, which will give us s squared minus four s plus four, which we can factor into s minus two most squared. So, therefore are the Austrians were X one will be asked, minus six over s minus two squared. We're gonna do partial fraction decomposition on this. So a fraction for s minus two and a fraction for s minus two. All squared. We're gonna most my through by the denominator to get that s minus six. That's the vehicle to a times X minus two plus B foiling out all the terms will get that a s minus to a plus B as a vehicle to s minus six. So with the magical efficient, therefore has people the one and negative to a which is just one negative too. Plus, B has to be equal to negative six and therefore being has your vehicle to negative four. Therefore our LaPlace transform of X one is that a which is one over s minus two minus four over s minus two, all squared. Never gonna take the inverse, the plaice transform of both sides to get that next won t is equal to the inverse a posturings former one of s minus two minus four turns the inverse of loss Transform of one over X minus two Well squared. And we can use the rule that they have lost transform of e to the 18 times Some function of T is able to of the class transport of that f at s minus A To write the first term as me to the to t times the universal class transform of one over s. And this it's just one since the LaPlace transform of tea to the end is an factorial over. As to the end plus one, we can write one over s as zero factorial over s 20 plus one. Therefore, the inverse of boss Transformers teach of the zero, which is just one minus four and we'll do the same thing over here. You need to the to t times the inverse applause transform, uh, one over x squared, which we can write as one factorial over s to the one plus one. So therefore, this his team say, for our X need to the to t fashioning out near the duty one minus 14. That's our X one. Now we're going to do the same thing for x two. How, uh, x two. The last transform of X two is equal to the determinant of being too. Over the determinate of a We're going to calculate the determinants of B two, which will be replacing the second row in a was being where our area shown here. So we'll have s one negative to one s one there too. And this determinant will come out to be s plus two says that the s plus two over s minus two. You all squared. We want to write. This is two fractions. A over s minus two must be over s minus two, All square. We're almost by through the denominator to get that s supposed to physical to a times asked Mines too. Plus being must be we're gonna boil up the term. Here's A S minus to a plus B is equal to s plus two. It's therefore means the match coefficients A is equal to one and negative two times a is one plus being is equal to two. Therefore, B is equal to floor. It's there for a distraction this one over X minus two plus four rs minus two mile square. So if we're taking the inverse of loss, transform our X two of tea will be the inverse applause transform of one over X minus two plus four terms and versus loss. Transform of one over ass minus two square. They were just figured out what these two are. So this will be do you to the negative or the positive two t times one plus four times t


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