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Assume that a Lyapunov function $V$ exists then $$\dot{V} = \Big(x\dfrac{\partial V}{\partial x}-y^{3}\dfrac{\partial V}{\partial y}\Big)(y^4-2x^2)-y\dfrac{\partial V}{\partial x}+2x\dfrac{\partial V}{\partial y}$$ This suggests the function $V = x^2+\dfrac{y^2}{2}$ which is postive definite and radially unbounded. Further, we have $$\dot{V} = -(y^4-2x^2)^2... 3 Using x_1 = x and x_2 = y, you have the nonlinear planar system$$ \begin{align} \dot{x}_1 &= x_1 x_2^4 - 2 x_1^3 - x_2 \\ \dot{x}_2 &= 2 x_1 + 2 x_1^2 x_2^3 - x_2^7 \end{align} \tag{1} $$which has an equilibirum at (x_1, x_2) = (0, 0). We can choose the Lyapunov function$$ V(x_1, x_2) = x_1^2 + \frac{1}{2} x_2^2 $$which is globally ... 1 Apologies, I would of preferred to try my best to finish the problem, but I'm just so busy. It's hard for me to picture the limit does not hold for a large class of well behaved functions. On the other hand surely we can contrive functions which will break the limit. By work such as Lebesgue's I feel finding a necessary and sufficient condition on functions ... 1 Once you have P(t)=Ce^{rt}, the statement that it doubles in five years means \frac {P(5)}{P(0)}=2. The value of C does not matter because it divides out. In fact, C=P(0). It doesn't matter whether P(0)=1 or 1000 or something else. Now you have$$2=e^{5r}\\ \log 2=5r\\r=\frac {\log 2}5$$which is not the same as r=\sqrt2. \sqrt 2 ... 1 If you take \alpha=4 you get the directional derivative equal zero. That means that V is not just some Lyapunov function but it is a first integral of your system. The orbits are level sets of such V which are circles. (0,0) is stable but not asymptotically stable. 1 Hint: Consider the general equation:$$ m\ddot{x}+c\dot{x}+kx =F_0(t)\tag1 $$and let x(\omega,t) be the particular solution of:$$m\ddot{x}+c\dot{x}+kx =e^{i\omega t}.\tag2$$Then the particular solution of (1) can be computed as:$$ x(t) = \int_{-\infty}^\infty F_0(\omega)x(\omega,t)d\omega, $$where$$ F_0(\omega)=\int_{-\infty}^\infty F_0(t)e^{-i\...

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Trying to write a short and easy to read answer: The amount of substance in water: $$x*V$$ The change rate of substance in water: $$(x*V)'$$ The change rate of water level: $$V' = s_1-s_2$$ The rate at which the substance is flowing into the tank is: $$0.1*s_1$$ The rate at which the substance is flowing out of the tank is: $$x*s_2$$ So you end up with ...

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You need to find the rate of change of alcohol in the tank. The rate is foun by rate in minus rate out Rate in is $$0.10s_1$$ Rate out is $$\frac {x(t)}{V(t)}s_2$$ Thus $$x'(t)=0.10s_1-\frac {x(t)}{V(t)}s_2$$ Assuming that the volume stays constant the equation is easier to solve.

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You are right that $v_1 = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$ is an eigenvector of $A$ with eigenvalue $2$; i.e $v_1 \in \ker(A-2I)$. But like you mentioned, the dimension of this kernel is $2$, so you need to find another linearly independent eigenvector. It is easy to verify that $\xi = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}$ is such a vector. ...

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Your eigenvectors are not right for $\lambda=2.$ You have the right system, but you should get two solutions: $v_1=(1,1,0)$ and $v_2=(1,0,1).$

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