Given,

`x^3y' + 2y = e^(1/x^2) ` and to find the particular solution of differential equation at `y(1) = e`.

so proceeding further , we get.

`x^3 y' + 2y = e^(1/x^2)`

=>`y' + 2y/(x^3) = e^(1/x^2) /x^3`

so , the equation is linear in y

and is of the form

`y' +p(x)y=q(x)`

so the general solution is given as

`y*(I.F)= int q(x) * I.F dx+c`

where I.F (integrating factor ) = `e^(int p(x) dx)`

on comparing we get ,

`p(x) = 2/x^3 and q(x) = e^(1/x^2) /x^3`

so ,

`I.F = e^(int (2/x^3) dx) = e^(2 (x^-3+1 )/ -2) = e^(-(x^-2))`

so ,

`y (e^(-(x^-2)))= int (e^(1/x^2) /x^3) * (e^(-(x^-2))) dx+c`

=>`y (e^(-(x^-2)))= int (x^-3) dx+c`

=>`y (e^(-(x^-2)))= x^((-3+1)/ -2)+c`

=> `y (e^(-(x^-2)))= x^-2/ -2+c`

=> `y = (- (x^-2)/2+c)/(e^(-(x^-2))) `

= `e^((x^-2)) *(c-(x^-2)/2 )`

so , now to find the particular soultion at `y(1) =e` , we have to do as follows

`y(x) = e^((x^-2)) *(c-(x^-2)/2 )`

=> y(1) = `e^((1^-2) ) *(c-(1^-2)/2 )`

=> `e= (e ) *(c-(1)/2 )`

=> `1= c-1/2`

=> `c= 3/2`

so the particular solution is

`y= ((e^((x^-2))) ) *(3/2-(x^-2)/2 )`

=`e^((x^-2)) *((3-(x^-2))/2 )`

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