Duckside
Reputation
438
Next privilege 500 Rep.
Access review queues
 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 $y^2 - 1 = y'$ Feb 17 asked Diff Eq. : Find an explicit solution of $y^2 - 1 = y'$ Jan 25 asked Differential Equation Basic - please explain the detail of this step Jan 25 comment Differential Equation: Using substitution $u(x)=y+x$, solve $\frac{dy}{dx} = (y+x)^2$ what if $u(x) = y^3$. what would be $\frac{dy}{dx}$ in this case. Jan 25 comment Differential Equation: Using substitution $u(x)=y+x$, solve $\frac{dy}{dx} = (y+x)^2$ thank you! I see i now Jan 25 reviewed Approve Differential Equation: Using substitution $u(x)=y+x$, solve $\frac{dy}{dx} = (y+x)^2$ Jan 25 accepted Differential Equation $y' = \frac{ty(4-y)}{(1+y)}$