Question

Verify that: y=xe^x is a solution of y''-2y'+y=0. Solve the initial value problem: y(0)=7, y'(0)=4

Answer 1

\[y'=e^x+xe^x\]\[y''=e^x+e^x+xe^x = 2e^x+xe^x\]
substitute in the equation\[2e^x+xe^x-2(e^x+xe^x)+xe^x\]if it equals zero thhen its a solution

Answer 2

y=xe^x
differentiating w.r.t x
y'=xe^x+e^x
again differentiating w.r.t x
y''=xe^x+e^x+e^x
y''=xe^x+2e^x
now y'=xe^x+e^x
i.e y'=y+e^x
2e^x=2y'-2y putting this in y'' equation
y''=xe^x+2y'-2y
y''=y+2y'-2y
y''-2y'-y=0

Answer 3

y''=2e^x+xe^x
on integration we get
y'=2e^x+xe^x-e^x+c1 ( c1=constant)
y'=e^x+xe^x+c1
now integrating to get y
y=e^x+xe^x+c1x-e^x+c2
y=xe^x+c1x+c2
we have to find c1 and c2
y(1)=7
puttiong x=0 in y
7=c2
now y'(0)=4
putting x=0 in y'
4=1+c1
c1=3
so c1=3 and c2=7
y=xe^x+3x+7
y'=e^x+xe^x+3