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`y' + (2x-1)y = 0 , y(1) = 2` Find the particular solution of the differential equation that satisfies the initial condition

Given ` y'+(2x-1)y=0`


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'+(2x-1)y=0--------(1)`


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


on comparing both we get,


`p(x) = (2x-1) and q(x)=0`


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 (2x-1) dx) *(0)) dx +c)/e^(int (2x-1) dx)`


first we shall solve


`e^(int (2x-1) dx)=e^(x^2 -x) `     


so


proceeding further, we get


y(x) =`((int e^(int (2x-1) dx) *(0)) dx +c)/e^(int (2x-1) dx)`


= `((int e^(x^2 -x)  *(0)) dx +c)/(e^(x^2 -x) )`


=`0+c/e^(x^2 -x)` = `e^(x-x^2+c )  `


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


to find the particular differential equation we have


y(1)=2


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


=> `e^c =2`


=> `c = ln(2)`


`y(x) = e^(x-x^2+ln(2))` 


`y(x) = e^ln2e^(x-x^2)` 


So,  


`y(x) = 2e^(x-x^2)` 

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