Given` 2xy' - y = x^3 - x`

=>`y' -(1/2x)y = (x^2-1)/2`

when the first order linear ordinary differential equation has the form of

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

then the general solution is ,

`y(x) = (int (e^(int p(x) dx) * q(x) dx + c )) / e^(int p(x) dx)`

so,

`y' -y/(2x) = (x^2-1)/2--------(1)`

`y'+p(x)y=q(x)---------(2)`

on comparing both we get,

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

so on solving with the above general solution we get:

`y(x) = (int (e^(int p(x) dx) * q(x) dx + c )) / e^(int p(x) dx)`

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

first we shall solve

`e^(int -1/(2x) dx)=e^(-ln(2x)/2)=1/sqrt(2x) `

so

proceeding further, we get

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

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

`=(1/(2*sqrt(2))(int1/sqrt(x) *((x^2-1)) dx +c)/(1/sqrt(2x) ))`

=`(1/(2*sqrt(2))(int1/sqrt(x) *(x^2)dx-int1/sqrt(x) *1 dx +c)/(1/sqrt(2x) ))`

`=((1/(2*sqrt(2))(2x^(5/2)/5-2sqrt(x)) +c)/(1/sqrt(2x) ))`

now to find the particular solution of differential equation we have ` y(4)=2`

so we can find the value of c

`y(x)=((1/(2*sqrt(2))(2x^(5/2)/5-2sqrt(x)) +c)/(1/sqrt(2x) ))`

=>`((1/(2*sqrt(2))(2x^(5/2)/5-2sqrt(x)) +c)*(sqrt(2x) ))`

=>`((1/(2)(2x^(5/2)/5-2sqrt(x)) +c)*(sqrt(x) ))`

=>`((x^(5/2)/5-sqrt(x) +C)*(sqrt(x) ))`

=>`((x^(3)/5-(x) +C*sqrt(x))))`

=> `y(4) =4^(3)/5-(4) +C*sqrt(4)`

=> 2 = `64/5-(4) +C*sqrt(4)`

=>` 2 = 64/5 -4+2c`

=>` 6- 64/5 =2c`

=> `-34/5 =2c`

=> `c= -17/5`

y(x)=`x^(3)/5-(x) -17/5sqrt(x)`

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