Solve the initial value problems in Exercises 55–58.

Question

57. d²y/dx² = 2e^(−x),y(0) = 1,y′(0) = 0

Accepted Answer

Hello there. Today we are going to solve the following practice problem together. So, first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Solve the initial value problem D2 Y by D X 2 is equal to 4 multiplied by E to the power of -3 X. Y 0 is equal to 2, Y of 0 is equal to 0. OK. So it appears for this particular problem we're asked to solve the initial value problem based on the information that is provided to us by the problem itself. So now that we know what we're ultimately trying to solve for, our first step that we need to take is we need to integrate D2 or D2 Y by D X 2. Is equal to 4 multiplied by E to the power of. -3 X with respect to, I'm gonna denote as WRT with respect to X. So once again we need to integrate with respect to X, so we need to take Dy by DX. So we're going to take DY by DX is equal to the integral of 4 multiplied by E to the power of -3XDX. Which will be equal to drumroll, -4 divided by 3, multiplied by E to the power of -3 X plus C subscript 1, which I'll just call C1 for short. Our second step now, so the step 2, is we need to integrate Dy by DX is equal to -4 divided by 3, multiplied by E to the power of -3 X plus C1 with respect to X, fantastic. So we are once again integrating with respect to X. So when we do that, we will find that why. will be equal to the integral of parentheses 4 divided by 3 multiplied by e to the power of -3 X plus C1 and don't forget your parenthesis on the end, multiplied by DX. And this will be equal to 4 divided by 9, multiplied by E to the power of -3 X plus C1 multiplied by X plus C2. Fantastic. Our third step now is we need to recall and apply the initial condition, which, as we should recall, the initial condition is Y of 0 is equal to 0. So we're going to be plugging in a value of 0 in for X, so we will find or in for Y specifically, so for X, so we're plugging in a value of 0 in. and when we do, we will find that 0. is equal to or divided by 9 multiplied by e to the power of 0 + 0 + C2 C subscript 2, which will mean we can write that 0. Equal to 4 divided by 9 plus C2, which when we rearrange this expression to isolate and solve for C2, we will find that C2 is simply equal to -4 divided by 9. OK. So our 4th and so our 4th step, so step number 4, is we need to recall and apply the initial condition, which once again, the initial condition is Y 0 is equal to 2. So we can go ahead and write that Y of X will be equal to -4 divided by 3, multiplied by E to the power of -3 X plus C1. So that will mean we can go ahead and write that 2 is equal to -4 divided by 3, multiplied by E to the power of 0 plus C1. Which will mean you go ahead and write that 2 is equal to -4 divided by 3 plus C1, which will mean when we rearrange this expression to isolate and solve for C1 using a little algebra, we will find that C1 is equal to 2 + 4 divided by 3, which will be simply equal to 10 divided by 3. So our fifth and final step, so this is our last step, is we need to write our final solution, which our final solution will be that Y of X is equal to 4 divided by 9 multiplied by E to the power of -3 X. Plus 10 divided by 3 X minus 4 divided by 9. And this is our final answer. Hooray, we did it. So going back to look at the problem itself to recap what we've learned, our final answer without a doubt once again is. That Y of X, so our function Y of X is equal to 4 divided by 9 multiplied by e to the power of -3 X plus 10 divided by 3X minus 4 divided by 9. And that's it. We've officially solved for this problem. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Solve the initial value problems in Exercises 55–58.57. d²y/dx² = 2e^(−x),y(0) = 1,y′(0) = 0

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Solve the initial value problems in Exercises 55–58.

57. d²y/dx² = 2e^(−x),y(0) = 1,y′(0) = 0