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Verify that the given function is a particular solution to the specified nonhomogeneous equation. Find the general solution and evaluate its arbitrary constants to find the unique solution satisfying the equation and the given initial conditions.$$y^{\prime \prime}+y=x, \quad y_{\mathrm{p}}=2 \sin x+x, \quad y(0)=0, y^{\prime}(0)=0$$

Calculus 3

Chapter 17

Second-Order Differential Equations

Section 2

Nonhomogeneous Linear Equations

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So we have the following question: y double prime. Following second order differential equation y fine plus y is equal to x. We are given a particular solution were given that some solution to this equation is 2 sine of x plus 6 point, and so you want to find the general solution. Why? Satisfying the conditions that are y prime 0 is equal to 0 and, while at 0 is 0 is equal to 0 tore. His particular solution satisfies this equation with the result on an homogeneous part and then, while we can figure out what is that by the nectar determined undetermined coefficient. So we assume that this is a combination of functions if there are x, so some combination of that. So the homogeneous is contest 0, so you'd have for these coefficients. We have r squared plus 1 is equal to 0. So this we know all that is going to be solution. Be r square is equal to minus 1. It means that you have complex coefficients there. So i can plus or minus in so the homogeneous is on be a combination of cosine and sine functions, and so we add to this particular. The general solution is going to be the homogeneous plus the particular solution, so that it will be some a cos of x, plus, b sine of x, plus this plus 2 sine of x plus x, and these 2 comes with than with them together so b, plus 2 and the sine of x unto you can call this new varia t some constant so that in general the solution is 1 of the a cos cos x, plus 3 sine x plus x. We want to satisfy these conditions of that's gonna. Give us the values of these 2 concerts, so we have to impose the condition that the function- a series equal to 0, so the function that 0 will be. We evaluate that will be a cos t 0 plus c sine 0 plus 0 pont. So all these sine of 0 is 0, so 0 and cost 0 is 1, so that is gone go to a so that is, sainete was 0, and now what is y prime? Well, since we know that a is going to be, serisapart is not going to be in the solution in the solution for that question, satisfying those those conditions otani, we differentiate these we get minus b, cos minus. We differentiate this minus c c x, not minus b, but minus minus cos of x and then differentiate x. That is 1 and all the regret will be. The expression of x equals to 0 to x, equals to 0 cos of 0, which is 1. So this will turn out to be minus c plus 1, and that has to be equal to 0. So that is saying that a wife move over there will get that 1 most be equal to z, and so our general solution is going to be this. With t equals to 1 point, so y is going to be a sine of x, plus x, or you can see that it satisfies those conditions, because y prime at 0 will be sine of 0 plus 0, which is 0 and then y prime o, o o E, a wee tibi only stay here, so the time sign c sign the eating. This is c cosine, so we have this equation. This would not be a minus and then, with that change we have. That c is equal to have this, and then we can move the 1 over there to obtain. That c has to be equal to minus 1, and then we have these minus sine y plus x. So that is important that that is. Why is important to check so you know this yeris 0 plus 0 and y prime will be minus cross y prime 101, and so we evaluate that at 0 would be minus cos, 0 plus 1, which is equal to 1 minus 1 plus 1. It is 0 point so so the solution of the general solution will be all the solutions of ifying. Those conditions solution to the differential equation, satisfying those conditions, y prating at z, is equal to 0, and yet is equal to 0 will be, will be this.

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