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visits member for 1 year, 9 months
seen May 5 at 17:56

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
28
revised Why does the general solution to $y'' + 4y = 0 $ contains sin and cos
edited title
Feb
28
comment Why does the general solution to $y'' + 4y = 0 $ contains sin and cos
thank you . Why don't you post this as an answer and I'll mark it as correct.
Feb
28
asked Why does the general solution to $y'' + 4y = 0 $ contains sin and cos
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?
Feb
17
comment Differential Equation $ (2x^2 + y^2)\,dx - xy \, dy = 0 $
Now I have,$$y' = \frac{2x^2 + y^2}{xy} $$ $$ yy' = 2x + \frac{y^2}{x}$$
Feb
17
asked Differential Equation $ (2x^2 + y^2)\,dx - xy \, dy = 0 $
Feb
17
accepted Diff Eq. : Find an explicit solution of $y^2 - 1 = y'$