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Consider the following initial-value problem_ Y 2y = 0, y(0) 1,Y(0) 2 Solve the given problem first using the form of the general solution given in (10) in Section ...

Question

Consider the following initial-value problem_ Y 2y = 0, y(0) 1,Y(0) 2 Solve the given problem first using the form of the general solution given in (10) in Section 4.3. Y = Ciekx + ~kx Cze (10) 2 +1 V2x + 1-V2 CV2x y(x) e 2 2Solve again, this time using the form given in (11) in Section 4.3_ y = C1 cosh kx + C2 sinh kx (11)y(x)

Consider the following initial-value problem_ Y 2y = 0, y(0) 1,Y(0) 2 Solve the given problem first using the form of the general solution given in (10) in Section 4.3. Y = Ciekx + ~kx Cze (10) 2 +1 V2x + 1-V2 CV2x y(x) e 2 2 Solve again, this time using the form given in (11) in Section 4.3_ y = C1 cosh kx + C2 sinh kx (11) y(x)



Answers

Find the particular solution of the differential equation that satisfies the boundary condition. $$ \begin{array}{ll} \underline{\text { Function }} & \underline{\text { Differential Equation }} \\ x d y=(x+y+2) d x &\quad y(1)=10 \end{array} $$

Hello, everyone. Today we're going to start from the number 23 here. The given first order differential equation is X divide equals X plus right plus two. The X so and wondered condition is going like buy off. Barnicle's took 10 so divide by the X equals X plus y plus two okay by X delayed by the X minus one by X in the way it was. Eggs plus two Divided by X. This p affects medicine bags, and this is cure Fixed X plus two by X So integrating factor equals e integral minus one by X the X, which is equal to erase toe minus Ln X, which is equals toe access to minus or which is equal to one by x. So why in tow one by X equals integral one by X into X plus To buy eggs The X so y equals x integral one by X plus to express to minus two, the X y equals X at Lenox minus two plus C x Boundary condition is way off. Barnicle's took 10 so 10 equals one in tow. Ellen one minus two plus e off right, 10 nickels C minus two c equals to 12. So why equals eggs? Atlantic plus two elects minus two. Thank you

Okay, so we're by the time it's worth trying was 11. If you put this up, let's rewrite this, uh, as a characteristic equation to get around the squared minus lambda plus 11 to 0. Okay, so it's factory us using our quarter equation. So it's negative or positive. That's one plus bonuses group of maybe 43. That's 43 I over to. So this is equal to 1/2. What's minus square Route 43. I over two. That's what. This sequel, Alfa this feed beta. So our solution since we have a complex conjugate Broots, we're gonna have e to our outfit, which is 1/2 times. See one co sign of our beta grew out of 43 over two x plus C to sign of. There were 43 over two. Thanks

In this question. We are given the differential equation why prime plus two X -1. Why Is equal to 0? And the boundary condition, Why is he going to two annexes? Go to one. Now first we look for our integrating factor μ which is given by E raised to the integral of the coefficient of Y, which is two X -1. D x mm. This is two X squared over the new power to minus X. And ignoring the constant for integration There we have eat of X squared minus six. Is our integrating factor. Not applying everything you know on our differential equation. By the integrating factor we have E to the power of X squared minus six. Why prime plus two X -1. Apply by E to the power of X squared minus X. And why Is equal to zero. And we recognize that the left hand side is equivalent to the derivative of each puff x squared minus X. By why? And this is equal to zero. Now integrating both sides with respect to X. We get e to the power of x squared minus x. Why being equal to see a constant? And writing both sides by E. To the power of x squared minus six. Get see over E to the X squared minus six which can also be read it and see E. Since we want to play the power bi minus one, we get x minus x squared. And this is the general solution. Now plugging in our initial value or boundary value we get since we have to hear and our c E raised to the power one minus one squared is equal to C. E. Races 40 is equal to see, so our C is equal to two. Therefore the final solution is the particular solution will be called to two over. Or since we had, we are now using this definition, so we have two mm to E to the power off X minus X squared. And this is our final particular solution to the given differential equation.

In this question. We are given the differential equation why prime plus two X -1. Why Is equal to 0? And the boundary condition, Why is he going to two annexes? Go to one. Now first we look for our integrating factor μ which is given by E raised to the integral of the coefficient of Y, which is two X -1. D x mm. This is two X squared over the new power to minus X. And ignoring the constant for integration There we have eat of X squared minus six. Is our integrating factor. Not applying everything you know on our differential equation. By the integrating factor we have E to the power of X squared minus six. Why prime plus two X -1. Apply by E to the power of X squared minus X. And why Is equal to zero. And we recognize that the left hand side is equivalent to the derivative of each puff x squared minus X. By why? And this is equal to zero. Now integrating both sides with respect to X. We get e to the power of x squared minus x. Why being equal to see a constant? And writing both sides by E. To the power of x squared minus six. Get see over E to the X squared minus six which can also be read it and see E. Since we want to play the power bi minus one, we get x minus x squared. And this is the general solution. Now plugging in our initial value or boundary value we get since we have to hear and our c E raised to the power one minus one squared is equal to C. E. Races 40 is equal to see, so our C is equal to two. Therefore the final solution is the particular solution will be called to two over. Or since we had, we are now using this definition, so we have two mm to E to the power off X minus X squared. And this is our final particular solution to the given differential equation.


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