Diferencial equations (lineal)?

Question

hi. i cant get to the book´s answer

ecuation: xy' + y = e^x initial value ----------------> y(1) = 2

book´s answer: y = x^-1 * e^x + (2-e)x^-1

Answer 1

there is a general form u need to remmeber the proof for which i dont know!

y'+y(function1 of x)=function2 of x  -------> the solution for this form is

y*e^ (integral of function1 of x)= integral of [(e^integral 0f function1 of x)*(function 2)]

plz try to understand the above going by the above

in ur question

y'+y/x=e^x/x

here 1/x=function1, e^x/x=function 2

=>y*e^(integral of 1/x)=integral of (e^integral of 1/x) *(e^x/x)

=>y*e^(log x) = integral of (e^log x)(e^x/x)

=>y*x=integral of(xe^x/x)  [e^logx =x]

=>xy=integral of e^x

=>xy=e^x+C

=>y=e^x/x  +C/x

when y=2 , x=1 (given!)

substisute to get C=2-e

therefore y=e^x/x + (2-e)/x

Answer 2

Write the equation in the form y - e^x = xy' = 0.  The equation is exact since d(y - e^x)/dy = 1 = d(x)/dx.  Now turn the crank for solving exact equations.