Find a particular solution yp of the following equation....

Find a particular solution $y_p$ of the following equation. Primes denote the derivatives with respect to x.\ y^(5) + 2y^(3) + 2y' = 8x^2 - 9\ The particular solution is $y_p(x) = $

Transcript

00:01 In this question, we need to find out the particular solution which is represented by yp of x of the following equation, which is y, fifth derivative of y plus two times the third derivative of y plus 2y double dash is equals to 2x square minus 8.

00:24 Okay so let us see how we are going to do this so the particular solution hence the particular solution which we will denote by y p of x it is equal to 1 over f of d times 2x square minus 8 correct so this is equals to 1 over.

00:51 Here i will get d -d -2 -5 plus 2d -qub plus 2d -square.

00:58 Here i will write that d is nothing but y -1.

01:03 That is y -d -d -s.

01:04 Okay.

01:05 So this is what i will get.

01:07 And here i'm having 2x square minus 8.

01:10 So now what i will do? i will take 2d -square common.

01:14 So i get 1 plus d -rase to 5 plus.

01:18 2dq divided by 2d square and here we have this function 2x square minus 8 this can also written as 1 over 2d square and this will i will write it as 1 over d cubed divided by 2 plus d in numerator it becomes raised to minus 1 and here we are having 2x square minus 8 now what else we can do here so this is nothing but 1 over 2d square and you can apply 1 plus x raised to minus 1 formula.

01:57 Okay, so 1 plus x raise to minus 1 is what? 1 minus x plus x square and so on.

02:04 Okay, so this formula we are using here.

02:07 So in place of x we have d cube over 2 plus d...

Question

Find a particular solution $y_p$ of the following equation. Primes denote the derivatives with respect to x.\ y^(5) + 2y^(3) + 2y' = 8x^2 - 9\ The particular solution is $y_p(x) = $

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