Question

(d^2y/dx^2)-5(dy/dx)+6y=xe^x

Answer 1

i know that y_h(t)=c_1e^(3t)+c_2e^(2t) but now i have to find y_p(t) so to do this I said it equals Axe^x. therefore y'=A(x+1)e^x and y"=A(x+2)e^x and i plug those back into the original equation and try to solve for A but am getting stuck.

Answer 2

This is a second order, linear, ordinary differential equation

Answer 3

u can use P.I(proper integral) and C.F(complementary function) to solve this

Answer 4

LHS can be written as\[(D^2-5D+6)y=0\]
so the auxiliary eq is--
\[m^2-5m+6=0\\(m-3)(m-2)=0\\m=2,3\\CF=c_1e^{2x}+c_2e^{3x}\]

Answer 5

now use this property--
\[\frac{ 1 }{ f(D) }e^{ax}V=e^{ax}\frac{ 1 }{ f(D+a) }V\]

Answer 6

then proper integral will be simply--
\[PI=\frac{ 1 }{ D^2-5D+6 }(x*e^{x})\\=e^{x}*\frac{ 1 }{ (D+1)^2-5(D+1)+6 }x\]
now try to simplyfy it

Answer 7

\[y''-5y'+6y=xe^x\]
With \[y=Axe^x\]\[y' = A(1+x)e^x\]\[y'' = A(2+x)e^x\]
Plugging these into the original DE\[A(2+x)e^x-5A(1+x)e^x+6Axe^x=xe^x\]
Factoring this equation\[A\Big((2+x)-5(1+x)+6x\Big)e^x=(x)e^x\]
\(e^x\) is not going to equal zero, so we can divide it out of the equation
\[A\Big((2+x)-5(1+x)+6x\Big)=x\\
A\Big(2x-3\Big)=x\]
Hmmm...

Answer 8

How did you get \(y_p(x) = Axe^x\)?

try again with \(y_p=(Ax+B)e^x\)

the derivatives are sure to be fun

Answer 9

I get A = 1/2 and B = 3/4
would anyone like to check?

Answer 10

Thank you guys so much! It looks like I had the wrong set up for the y_p(t) equation! Haven't actually gone through the problem but in the answer key A=1/2 and B=3/4 like alekos said

Answer 11

thats it!