2016年10月13日 · Think of dy and dx each as discrete variables. So you could do something like multiply both sides by dx and end up with: iff dy=ydx And then divide both sides by y: iff dy/y=dx Now, integrate the left-hand side dy and the right-hand side dx: iff int 1/y dy=int dx iff ln |y|=x+C Remember to add the constant of integration, but we only need one.

2023年2月27日 · This is a standard example of separation of variables. First note that the zeros of the expression y(1-y) can give us constant solutions, depending on the value of y 0. If y=1 (constant), then dy/dt = 0 = ky(1-y), because 1-y=0 in this case. Similarly if y=0 (constant), then dy/dt = 0 = ky(1-y).

2018年4月4日 · The General Solution is: y = 1-1/(e^t + C) We have: dy/dt = e^t(y-1)^2 We can collect terms for similar variables: 1/(y-1)^2 \ dy/dt = e^t Which is a separable First Order Ordinary non-linear Differential Equation, so we can "separate the variables" to get: int \ 1/(y-1)^2 \ dy = int e^t \ dt Both integrals are those of standard functions, so …

2021年5月27日 · $\tiny{b.1.2.2a}$ $\dfrac{dy}{dt}=y-5, \quad y(0)=y_0$$\begin{array}{rll}\displaystyle y'&=y-5\\ y'-y&=-5\\ u(x)&=\exp\int-1\, dx&= e^{-t}\\...

2009年1月12日 · Homework Statement Find the general solution of the differential equation specified. \\frac{dy}{dt} = ty. The Attempt at a Solution I already know the answer to be ke^{t^{2}/2}, but can't figure out how it got here. I'm rusty with my integrals and am just really starting diff eqs...

2010年5月2日 · Homework Statement If dy/dt = ky and k is a nonzero constant, than y could be a. 2e^kty b. 2e^kt c. e^kt + 3 d. kty + 5 e. .5ky^2 + .5 Correct answer is b. 2e^kt Homework Equations The Attempt at a Solution I don't really know what I should be doing. I have a feeling I...

2008年3月4日 · First look at dy/dt= -y. That's easy to solve and gives the general solution to the associated "homogeneous equation". Now, try something of the form y= At+ B, for constants A and B. Put that into the equation and try to find A and B that will make the equation true.

2014年9月21日 · There are two values depending on the point. {dx}/{dt}={(-8 " at " (3,4)),(8 " at "(-3,4)):} Let us look at some details. First, let us find the values of x. By plugging in y=4 into x^2+y^2=25, x^2+16=25 Rightarrow x^2=9 Rightarrow x=pm3 Now, let us find some derivatives. By differentiating with respect to t, d/{dt}(x^2+y^2)=d/{dt}(25) Rightarrow 2x{dx}/{dt}+2y{dy}/{dt}=0 by dividing by 2x ...

2005年2月4日 · because dy = dt, then you will get a graph where a small change in y (within the y axis) will give an EQUIVALENT small change in t (in the t axis) i believe it is the same as saying y = t, which is just a unit of y compared with a unit of t. …

2016年9月8日 · #(dy_h)/(dt)+3t y_h = 0# #(dy_p)/(dt)+3t y_p = sint# The homogeneous solution can be easily obtained. So grouping variables. #dy_h/y_h=-3tdt# giving. #y_h = C e^(-3/2t^2)# The particular solution can be obtained adopting the so called constants variation technique due to Lagrange. Posing. #y_p = C(t)e^(-3/2t^2)# and substituting in the ...