DramaQueen001
contestada

I have [tex]y = x^{2} + \frac{c}{ x^{2} } [/tex] and for any value of c, a solution of the equation is [tex]xy' + 2y = 4 x^{2} , (x\ \textgreater \ 0)
[/tex] and I am supposed to find the value of c for which y(3)=1. What I did was I took the differential of the first equation to get y' and substituted it in the second equation, along with the value for x (which is 3), to solve for c but when I put the value I got for c and the value of x into the first equation, I don't get y = 1. Could anyone please help me see what I'm doing wrong?

Respuesta :

LammettHash
Given [tex]y(3)=1[/tex], the solution gives the equation

[tex]1=3^2+\dfrac c{3^2}\implies1=9+\dfrac c9\implies c=-72[/tex]

so that the particular solution is

[tex]y=x^2-\dfrac{72}{x^2}[/tex]

To verify that this solution is correct, differentiate it, then plug it and its derivative into the ODE and arrive at an identity.

[tex]y'=2x+\dfrac{144}{x^3}[/tex]
[tex]\implies xy'=2x^2+\dfrac{144}{x^2}[/tex]

[tex]xy'+2y=4x^2\iff \left(2x^2+\dfrac{144}{x^2}\right)+2\left(x^2-\dfrac{72}{x^2}\right)=4x^2\iff 4x^2=4x^2[/tex]

which is true for all [tex]x>0[/tex].

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