- #1
Punchlinegirl
- 224
- 0
Use reduction of order to find a second solution to the given differential equation
(x-1)y"-xy'+y=0 x>1 y_1=e^x
Putting it in standard form gives
[tex] y"-x/x-1 y' + 1/x-1 y =0 [/tex]
y(t)=v(t)e^x
y'(t)= v'(t)e^x +v(t)e^x
y"(t)= v"(t)+2v'(t)e^x +v(t)e^x
plugging into the initial equation:
[tex] v"(t)e^x+2v'(t)e^x+v(t)e^x-xv'(t)e^x/x-1 +xv(t)e^x/x-1 +v(t)e^x/x-1 [/tex]
I'm not sure how to simplify this further if it can even be done, or what I should do next. Can someone please help me out?
(x-1)y"-xy'+y=0 x>1 y_1=e^x
Putting it in standard form gives
[tex] y"-x/x-1 y' + 1/x-1 y =0 [/tex]
y(t)=v(t)e^x
y'(t)= v'(t)e^x +v(t)e^x
y"(t)= v"(t)+2v'(t)e^x +v(t)e^x
plugging into the initial equation:
[tex] v"(t)e^x+2v'(t)e^x+v(t)e^x-xv'(t)e^x/x-1 +xv(t)e^x/x-1 +v(t)e^x/x-1 [/tex]
I'm not sure how to simplify this further if it can even be done, or what I should do next. Can someone please help me out?