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Just use the initial condition y(1)=sqrt(2). Hence, x=1 and y=sqrt(2) and you substitute these values into the solution. Then you have an equation with only one unknown "c".
Looks like you went wrong when you started integrating by parts. There is no need to integrate by parts at that point. The integral there is a simple ∫exp(u)du type.
Also, I don't think the answer you gave as the correct answer is correct. The quick way to check is to substitute that solution into the original diff. eq. and then check the initial condition. Actually, the proposed solution does not even match the initial condition.
Although integration by parts is a valid method, you made a mistake when you used it. In this case integration by parts leads to an unsolvable integral of the type ∫exp(x^2) dx.
It looks like I was using a wrong formula and on the other hand integration by parts technique fails in this case.
Regards
PG
PS: After reviewing it, I don't think choice of the formula has anything to do here because using any of the formulae leads to same result. So, where did I make a mistake? Kindly let me know. Thanks.
What on earth are you doing here. There are so many mistakes in integration in what you wrote. ???
Whenever you do an integral, it's a good idea to check it by taking the derivative of the answer and making sure you get back to where you started. In this way, you will find errors quickly. The bottom line is that ∫exp(x^2)dx doesn't have a closed form integral formula. Whichever of the two ways you do the integration by parts, you end up with that integral. Hence, ... dead end.
The integral ∫exp(x^2)dx can be related to the error function (erf), which is based on the integral of a Gaussian type function exp(-x^2). Both, ∫exp(x^2)dx and ∫exp(-x^2)dx are similar and require numerical methods or tables to find the values. But, this is information that is useless in the problem you are solving.
Note, that you should memorize the fact that these integrals are unsolvable, but are often expressed as the erf function. These integrals show up very often.
Does the same goes for the integral ∫x*exp(x^2)dx? Is it also not solvable? If your answer is 'yes' then this would also mean that the final answer reached for is incorrect. Thank you.
Of course ∫x*exp(x^2)dx is solvable. You solved it, and I said it was of type ∫exp(u) du, which is well known. You can solve it and take the derivative to prove it works, and you can take the answer to the problem and plug it in to the original differential equation and also check the initial condition.
The solution will be y= -C1+C2*ln(x). C1 can be any constant value that will swallow the sign, so what difference does the sign make? The expressions y=C1+C2*ln(x), y=-5+C2*ln(x), or y=5+C2*ln(x) all are solutions to the DE.
Q1: What you said seems correct, but only for "linear" diff. eqs. It's important to put that qualifier into your statement because it is not generally true for nonlinear cases.
Q2: No, the author is correct. You don't need to include all "n" possible solutions into the principle of superposition. The index "k" can be any integer between 1 and "n", inclusive.
Q3: I think what you say is essentially correct, but one part may not be quite right. You say there is only one particular solution, but my recollection is that this is not true. You are only required to find one particular solution to assemble the general solution to the nonhomogeneous equation. There may be more particular solutions (in fact, there must be), but you don't care about that because once you combine any particular solution with the "n" homogeneous solutions, you have the complete general solution.
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