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seen Oct 18 at 7:39

Apr
3
comment Why is the particular solution of $y'' - 4y' +3y = e^t$ not in the form of $Ae^t$
"both $e^t$ and $te^t$" meaning the homogenous equation looks like $Ay'' - By' + Cy = e^t + te^t$, then the particular solution will look like $t^2e^t$, am I correct?
Apr
3
comment Why is the particular solution of $y'' - 4y' +3y = e^t$ not in the form of $Ae^t$
and if $te^t$ is already in the homogeneous solution, we will need to multiply by one more $t$ making it $t^2e^t$?
Apr
3
asked Why is the particular solution of $y'' - 4y' +3y = e^t$ not in the form of $Ae^t$
Apr
1
accepted Particular Solution of $y'' - 3y' - 4y = 3e^{2t}$
Apr
1
comment Particular Solution of $y'' - 3y' - 4y = 3e^{2t}$
Thank you. I feel ashamed that I did not see this.
Apr
1
asked Particular Solution of $y'' - 3y' - 4y = 3e^{2t}$
Mar
31
accepted Differential Equation $y'' + 8y' - 9y = 0$ and Wronskian
Mar
31
comment Differential Equation $y'' + 8y' - 9y = 0$ and Wronskian
yes. thank you.
Mar
31
asked Differential Equation $y'' + 8y' - 9y = 0$ and Wronskian
Feb
18
accepted Differential Equation, finding where the solution is bounded
Feb
18
revised Differential Equation, finding where the solution is bounded
added 24 characters in body
Feb
18
asked Differential Equation, finding where the solution is bounded
Feb
18
accepted What is the particular solution of $y'' - 4y' + 3y = 2t + e^t$
Feb
18
asked What is the particular solution of $y'' - 4y' + 3y = 2t + e^t$
Feb
18
comment Differential Equation $y'' - 4y' + 4y = 0$
Then in what case will it has a form of $Ce^tcos(t)...Ce^tsin(t) $
Feb
18
accepted Differential Equation $y'' - 4y' + 4y = 0$
Feb
17
comment Differential Equation $y'' - 4y' + 4y = 0$
is it considered to have "repeated roots" if the roots are (r-2)(r+2)? (roots of 2 and -2). No right?
Feb
17
asked Differential Equation $y'' - 4y' + 4y = 0$
Feb
17
accepted Differential Equation $ (2x^2 + y^2)\,dx - xy \, dy = 0 $
Feb
17
comment Differential Equation $ (2x^2 + y^2)\,dx - xy \, dy = 0 $
In most problem (that I've seen so far) that are homogeneous, you set $ u =\frac{y}{x}$ , if I set $ u =\frac{x}{y}$ would that be a problem or would it just give you the same thing?