Question

Find the general solution of x^2*y"+xy'-y=4/x^2, given that y1=x satisfies the complementary equation.

Answer 1

\[x ^{2}y''+xy'-y=\frac{ 4 }{ x^2 }\]

Answer 2

y=ux
y'=u'x+u
y"=u"x+2u'
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Subbing:
\[x^2y''+xy'-y=x^2(u''x+2u')+x(u'x+u)-ux=\frac{ 4 }{ x^2 }\]

Answer 3

Simplify, I got
\[x^3u''+3x^2u'=\frac{ 4 }{ x^2 }\]

Answer 4

Substitution: z=u'
x^3*z'+3x^2*z=4/x^2
z'+(3/x)z=4/x^5
integrating factor is x^3
so d/dx(x^3*z)=4/x^2
x^3*z=integral of 4/x^2 dx
x^3*z=-4/x+C
so \[z=-\frac{ 4 }{ x^4 }+\frac{ C }{ x^3 }\]

Answer 5

Since z=u',
u'=-4/x^4+C/x^3
integrate u',
\[u=\frac{ 4 }{ 3x^3 }-\frac{ C _{2} }{ 2x^2 }+C _{1}\]

Answer 6

\[y=\frac{ 4 }{ 3x^2 }+C _{1}x+\frac{ C _{2} }{ 2x }\]

Answer 7

But the book's answer is y=4/3x^2+C1*x+C2/x. So I don't know where I got wrong. Please tell me where my mistake is.

Answer 8

@Directrix @pooja195 @surjithayer @Nnesha @Preetha

Answer 9

@Kainui

Answer 10

They look the same to me, since \(\frac{C_2}{2}\) is just as arbitrary of a constant as plain \(C_2\) they just got rid of the 2 because it doesn't change the answer.

Answer 11

So you're saying that \[\frac{ C _{2} }{ 2x }=\frac{ C _{2} }{ x }\]?

Answer 12

And integrating C/x^3 gives you -C/2x^2 but it's also the same thing as C/2x^2, right? Since you said the term got absorbed into the constant C, right? So the book's answer is correct?

Answer 13

Yeah, the book is right and you are right. One thing though, actually using the same letter on both sides makes it false, because if you were to solve that
\[\frac{C_2}{2x} = \frac{C_2}{x}\] you'd get that:

\[C_2 = 2*C_2\]

which implies \(C_2=0\) which is not what you mean. So it's technically correct to write something to distinguish them apart, like use a different letter.

\[\frac{C_2}{2x} = \frac{D_2}{x}\]

This is fine, cause all this says is \(C_2 = 2* D_2\). But no arbitrary constant is better than any other arbitrary constant so w/e

Answer 14

Okay, thank you so much for pointing out. :)