Question

Use variation of parameters to find a particular solution of x^2*y"+xy'-y=2x^2+2, given that y1=x and y2=1/x satisfy the complementary equation.

Answer 1

\[y _{p}=u _{1}x+u _{2}\frac{ 1 }{ x }\]
\[u'_{1}x+u'_{2}\frac{ 1 }{ x }=0\]
\[u'_{1}-u'_{2}\frac{ 1 }{ x^2 }=2x^2+2\]
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\[2u'_{2}\frac{ 1 }{ x }=-2x^3-2x\]
\[u'_{2}=-x^4-x^2\]
\[u _{2}=-\frac{ x^5 }{ 5 }-\frac{ x^3 }{ 3}\]
\[u'_{1}=x^2+1\]
\[u _{1}=\frac{ x^3 }{ 3 }+x\]
\[y _{p}=\frac{ x^4 }{ 3 }+x^2-\frac{ x^4 }{ 5 }-\frac{ x^2 }{ 3 }\]
But the answer in the book says \[y _{p}=\frac{ 2(x^2-3) }{ 3 }\]. What's wrong?

Answer 2

@freckles

Answer 3

You forgot divided both sides by x^2, so that your f(t) ( the RHS) is not correct. It should be \(2+\dfrac{2}{x^2}\)

Answer 4

This mistake make your \(u_1'-u_2'\dfrac{1}{x^2}=2x^2+2\) becomes wrong.

Answer 5

you need to bring
x^2*y"+xy'-y=2x^2+2
in the form
\( y'' + p(x) y' + q(x) y = g(x)
\)
only then you'll able to use the method of parameters this way.

So, as suggested by Loser,
please divide that entire equation by x^2 first

Answer 6

\(u'_{1}-u'_{2}\frac{ 1 }{ x^2 }=2+\frac{2}{x^2} \)