- #1

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- Thread starter Fatima Hasan
- Start date

- #1

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- #2

Buzz Bloom

Gold Member

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I suggest you take your solution

y(x) = x^{2} - (1/3)x^{-1}

and calculate y'(x) and y"(x).Then substitute these three functions of x in to corresponding terms of the original equation. After the substitution, the resulting equation should be reducible to

0 = 0

if your and answer is correct.Regards,

Buzz

- #3

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I didn't get this equation , can you tell me where is my mistake please ?the resulting equation should be reducible to

0 = 0

- #4

Mark44

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Thread moved.

- #5

Mark44

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Your answer for ##y_2(x)## in the next-to-last line in post #1 is incorrect. Check the integration that you did.I didn't get this equation , can you tell me where is my mistake please ?

- #6

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##y_2(x) = x(x-\frac{1}{x})##Your answer for ##y_2(x)## in the next-to-last line in post #1 is incorrect. Check the integration that you did.

##y_2(x) = x^2-1##

##y(x) = C_1 y_1(x) + C_2y_2(x)##

##y(x) = C1 x + C_2 (x^2-1)##

- #7

Mark44

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OK, now can you check that your solution is correct?##y_2(x) = x(x-\frac{1}{x})##

##y_2(x) = x^2-1##

##y(x) = C_1 y_1(x) + C_2y_2(x)##

##y(x) = C1 x + C_2 (x^2-1)##

- #8

- 318

- 13

Got it , thank you.OK, now can you check that your solution is correct?

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