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

Expert Answers

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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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