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I am a high school student. I am having some problems with the following question and can't solve it. I need help to solve this. if $y$ = $x^2\cdot \cos x$

What will be the value of: $$(x^2)d^2y/dx^2 - (4x)\cdot dy/dx + (x^2+6)y $$

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Do you know what the product rule is? –  Vectk Apr 26 '13 at 8:12
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You mean , $d(uv)/dx$ = $v*(du/dx) + u*(dv/dx)$ Yes I do. I tried this but it is becoming too lengthy and finally the answer is not right –  shiladitya basu Apr 26 '13 at 8:15
    
No!. (uv)'=u'v+uv' @shiladityabasu. –  Babak S. Apr 26 '13 at 8:15
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OE means? @BabakS. –  shiladitya basu Apr 26 '13 at 8:19
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Yes I got it. I got confused about the product rule of differentiation. Yes I know the product rule and applied that on the expression $y = x^2*cos x$ and it became so lengthy and ended up in disaster. –  shiladitya basu Apr 26 '13 at 8:22

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I think this is likely to be more about organising your work than knowing the basic things. You want to take $u=x^2; v=\cos x$ in the product rule.

Differentiating $u$ will always give powers of $x$ or constant terms and differentiating $v$ will always give a term in sin or cos, and you will always have something derived from $u$ times something derived from $v$ - so you know your answer will be something of the form $p(x)\cos x + q(x)\sin x$.

Then you can note (using the product rule twice) that $(uv)''=u''v+2u'v'+uv''$

Then you need to make a decision whether you are going to organise terms first by the power of $x$ or alternatively by whether they have sin or cos. I have always found it helpful (if I have something a bit complicated) to take a line for each term so $$x^2y''=x^2(u''v+2u'v'+uv'')=$$

The second term goes on line 2 and the final one on line 3, with the final expressions arranged on the right hand side of the line in accordance with the scheme I have chosen. If I have done it right, I have like terms (eg all the terms in $x^2\cos x$) in columns on three lines on the right hand side and I can add them.

But if you get the calculations right, I think you will not have as many terms or as much complexity as you imagine. I've put these notes in case they help you to see that in the end.

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Thanks! It helped a lot! –  shiladitya basu Apr 26 '13 at 9:00

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