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`2xy' - y = x^3 - x , y(4) = 2` Find the particular solution of the differential equation that satisfies the initial condition

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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