1
Test 4 / Spring 2020 Wednesday Sitting Q4
« on: December 03, 2020, 12:52:15 AM »
For question b in Q4, could anyone explain how we get this?
Thanks in advance!
Thanks in advance!
Show Posts
This section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.
Pages: [1] 2
2
Test 2 / Possible typo in 2018 Version B Q3(b)
« on: October 28, 2020, 04:13:02 PM »
In 3b, harmonic conjugate function v(x, y) is $3 - 12x^2y^2 + 2x^42y^4+5y+C$, does anyone know where the constant $3$ comes from?
3
Chapter 1 / Re: how to solve question 19 in section 1.2 in the textbook
« on: September 26, 2020, 08:06:10 PM »
Expanding the restriction reveals.
$\lvert z - p\rvert = cx \Rightarrow \sqrt{(x-p)^2 + y ^ 2} = cx \Rightarrow x^2 - 2xp + p^2 + y^2 = c^2x^2, x > 0$
$\Rightarrow (1 - c^2)x^2 - 2xp + y^2 = 0$
We know the sign of the quadratic component is what determines the behaviour. Thus
$c \in (0,1) \Rightarrow (1 − c_{2}) > 0 \Rightarrow \frac{{x'}^2}{a^2} + y^2 = 1 \Rightarrow$ ellipse
$c = 0 \Rightarrow (1 − c_{2}) = 0 \Rightarrow x = \frac{y^2}{2p} \Rightarrow$ parabola
$c \in (1,\infty) \Rightarrow (1 − c_{2}) < 0 \Rightarrow -\frac{{x'}^2}{a^2} + y^2 = 1 \Rightarrow$ hyperbola
where x' is the appropriate translation of x.
$\lvert z - p\rvert = cx \Rightarrow \sqrt{(x-p)^2 + y ^ 2} = cx \Rightarrow x^2 - 2xp + p^2 + y^2 = c^2x^2, x > 0$
$\Rightarrow (1 - c^2)x^2 - 2xp + y^2 = 0$
We know the sign of the quadratic component is what determines the behaviour. Thus
$c \in (0,1) \Rightarrow (1 − c_{2}) > 0 \Rightarrow \frac{{x'}^2}{a^2} + y^2 = 1 \Rightarrow$ ellipse
$c = 0 \Rightarrow (1 − c_{2}) = 0 \Rightarrow x = \frac{y^2}{2p} \Rightarrow$ parabola
$c \in (1,\infty) \Rightarrow (1 − c_{2}) < 0 \Rightarrow -\frac{{x'}^2}{a^2} + y^2 = 1 \Rightarrow$ hyperbola
where x' is the appropriate translation of x.
4
Quiz-7 / Re: Q7-T0801
« on: March 30, 2018, 01:26:18 PM »
(a)The critical points are given by the solution set of the equations $$-2x-x-y-x(x^2+y^2) = 0$$ $$x-y+y(x^2+y^2) = 0$$ It is clear that the origin is a critical point. Solving the first equation for C , we find that $$y = \frac{-1 \pm \sqrt{1-8x^2-4x^4}}{2x}$$
Substitution of these relations into the second equation results in two equations of the form $f_{1}(x) = 0$ and $f_{2}(x) = 0$ . Plotting these functions, we note that only$f_{1}(x) = 0$ has real roots. ('It follows that the additional critical points are at (-0.33076,1.0924) and (0.33076,-1.0924)
(b,c) Given that $$F(x,y) = -2x-x-y-x(x^2+y^2)$$ $$G(x,y) = x-y+y(x^2+y^2)$$
the Jacobian matrix of the vector field is $$\begin{pmatrix} -2-3x^2-y^2 & -1-2xy \\ 1+2xy & -1+x^2+3y^2 \end{pmatrix}$$
with complex conjugate eigenvalues $r_{1,2} =((-3 \pm i \sqrt{3})/2) $ . Hence the critical point is a stable spiral, which is asymptotically stable. At the point (-0.33076,1.0924) , the coefficient matrix of the linearized system is $$J(-0.33076,1.0924) = \begin{pmatrix} -3.5216 & -0.27735 \\ 0.27735 & 2.6895 \end{pmatrix} $$
With eigenvalues $r_1 = -3.5092$ and $r_2 = 2.6771$ The eigenvalues are real, with opposite sign. Hence the critical point of the associated linear system is a saddle, which is unstable. Identical results hold for the point at (0.33076,-1.0924).
Attached is the part(d)
Substitution of these relations into the second equation results in two equations of the form $f_{1}(x) = 0$ and $f_{2}(x) = 0$ . Plotting these functions, we note that only$f_{1}(x) = 0$ has real roots. ('It follows that the additional critical points are at (-0.33076,1.0924) and (0.33076,-1.0924)
(b,c) Given that $$F(x,y) = -2x-x-y-x(x^2+y^2)$$ $$G(x,y) = x-y+y(x^2+y^2)$$
the Jacobian matrix of the vector field is $$\begin{pmatrix} -2-3x^2-y^2 & -1-2xy \\ 1+2xy & -1+x^2+3y^2 \end{pmatrix}$$
with complex conjugate eigenvalues $r_{1,2} =((-3 \pm i \sqrt{3})/2) $ . Hence the critical point is a stable spiral, which is asymptotically stable. At the point (-0.33076,1.0924) , the coefficient matrix of the linearized system is $$J(-0.33076,1.0924) = \begin{pmatrix} -3.5216 & -0.27735 \\ 0.27735 & 2.6895 \end{pmatrix} $$
With eigenvalues $r_1 = -3.5092$ and $r_2 = 2.6771$ The eigenvalues are real, with opposite sign. Hence the critical point of the associated linear system is a saddle, which is unstable. Identical results hold for the point at (0.33076,-1.0924).
Attached is the part(d)
5
Quiz-7 / Re: Q-T5101
« on: March 30, 2018, 01:24:57 PM »
(a)The critical points consist of the solution set of the equations. $$(2+x)(y-x) = 0$$ $$(4-x)(y+x) = 0$$, the only critical points are at (0,0), (4,4)and (-2,2).
(b,c) First note that $F(x) = (2+x)(y-x) $ ,$G(x,y) = (4-x)(y+x)$ . The Jacobian matrix of the vector field is $$J = \begin{pmatrix} -2-2x+y & 2+x \\ 4-2y-2x & 4-x \end{pmatrix}$$
At the origin, the coefficient matrix of the linearized system is $$J(0,0) = \begin{pmatrix} -2 & 2 \\ 4 & 4 \end{pmatrix}$$
with eigenvalues $r_1 = 1-\sqrt{17}$, $r_2 = 1+\sqrt{17}$ . The eigenvalues are real, with opposite sign. Hence the critical point is a saddle, which is unstable. At the equilibrium point (-2,2), the coefficient matrix of the linearized system is $$J(-2,2) = \begin{pmatrix} 4 & 0 \\ 6 & 6 \end{pmatrix}$$
with eigenvalues $r_1 = 4$ and $r_2 = 6$ . The eigenvalues are real, unequal and positive, hence the critical point is an unstable node. At the point (4,4), the coefficient matrix of the linearized system is $$J(4,4) = \begin{pmatrix} -6 & 6 \\ -8 & 0 \end{pmatrix}$$
with complex conjugate eigenvalues $r_1 = -3 + \sqrt{39}i$, $r_2 = -3 - \sqrt{39}i$, The critical point is a stable spiral, which is asymptotically stable.
Attached is the part (d)
(b,c) First note that $F(x) = (2+x)(y-x) $ ,$G(x,y) = (4-x)(y+x)$ . The Jacobian matrix of the vector field is $$J = \begin{pmatrix} -2-2x+y & 2+x \\ 4-2y-2x & 4-x \end{pmatrix}$$
At the origin, the coefficient matrix of the linearized system is $$J(0,0) = \begin{pmatrix} -2 & 2 \\ 4 & 4 \end{pmatrix}$$
with eigenvalues $r_1 = 1-\sqrt{17}$, $r_2 = 1+\sqrt{17}$ . The eigenvalues are real, with opposite sign. Hence the critical point is a saddle, which is unstable. At the equilibrium point (-2,2), the coefficient matrix of the linearized system is $$J(-2,2) = \begin{pmatrix} 4 & 0 \\ 6 & 6 \end{pmatrix}$$
with eigenvalues $r_1 = 4$ and $r_2 = 6$ . The eigenvalues are real, unequal and positive, hence the critical point is an unstable node. At the point (4,4), the coefficient matrix of the linearized system is $$J(4,4) = \begin{pmatrix} -6 & 6 \\ -8 & 0 \end{pmatrix}$$
with complex conjugate eigenvalues $r_1 = -3 + \sqrt{39}i$, $r_2 = -3 - \sqrt{39}i$, The critical point is a stable spiral, which is asymptotically stable.
Attached is the part (d)
6
Quiz-7 / Re: Q7-T0501
« on: March 30, 2018, 01:13:14 PM »
(a) The critical points are solutions of the equations,$1-y = 0$, $(x-y)(x+y) = 0$
The first equation requires that y= 1 . Based on the second equation, x =1/-1. Hence the critical points are (-1,1) and (1,1).
(b,c) $F(x,y) = 1-y$ and $G(x,y) = x^2-y^2 $. The Jacobian matrix of the vector field is $$J = \begin{pmatrix} 0 & -1 \\ 2x & 2y \end{pmatrix}$$
At the critical point $(-1,1)$, the coefficient matrix of the linearized system is $$J(-1,1) = \begin{pmatrix} 0 & -1 \\ -2 & -2 \end{pmatrix}$$
with eigenvalues $r_1 = -1-\sqrt{3}$ and $r_2 = -1+\sqrt{3} $ . The eigenvalues are real, with opposite sign. Hence the critical point is a saddle, which is unstable. At the equilibrium point (1,1), the coefficient matrix of the linearized system is $$J = \begin{pmatrix} 0 & -1 \\ 2 & -2 \end{pmatrix}$$ with complex conjugate eigenvalues $r_1 = -1+i$, $r_2 = -1-i$. The critical point is stable spiral, which is asymptotically stable.
Attached is the part(d)
The first equation requires that y= 1 . Based on the second equation, x =1/-1. Hence the critical points are (-1,1) and (1,1).
(b,c) $F(x,y) = 1-y$ and $G(x,y) = x^2-y^2 $. The Jacobian matrix of the vector field is $$J = \begin{pmatrix} 0 & -1 \\ 2x & 2y \end{pmatrix}$$
At the critical point $(-1,1)$, the coefficient matrix of the linearized system is $$J(-1,1) = \begin{pmatrix} 0 & -1 \\ -2 & -2 \end{pmatrix}$$
with eigenvalues $r_1 = -1-\sqrt{3}$ and $r_2 = -1+\sqrt{3} $ . The eigenvalues are real, with opposite sign. Hence the critical point is a saddle, which is unstable. At the equilibrium point (1,1), the coefficient matrix of the linearized system is $$J = \begin{pmatrix} 0 & -1 \\ 2 & -2 \end{pmatrix}$$ with complex conjugate eigenvalues $r_1 = -1+i$, $r_2 = -1-i$. The critical point is stable spiral, which is asymptotically stable.
Attached is the part(d)
7
Quiz-7 / Re: Q7-T0401
« on: March 30, 2018, 01:01:22 PM »
(a)The critical points are given by the solution set of the equations.$$1-xy = 0$$ $$x-y^3=0$$
After multiplying the second equation by y , it follows that $y=1/-1$ . Hence the critical points of the system are at (-1,1) and (-1,-1).
(b,c) Note that $F(x,y) = 1-xy$ and $G(x,y) = x-y^3$ . The Jacobian matrix of the vector field is $$ J = \begin{pmatrix} -y & -x \\ 1 & -3y^2 \end{pmatrix} $$
At the critical point (1,1), the coefficient matrix of the linearized system is$$ J(1,1) = \begin{pmatrix} -1 & -1 \\ 1 & -3 \end{pmatrix}$$
with eigenvalues $r_1 = r_2 = -2$ . The eigenvalues are real and equal. It is easy to show that there is only one linearly independent eigenvector. Hence the critical point is a stable improper node.
At the point (-1,-1), the coefficient matrix of the linearized system is $$ J(-1,-1) = \begin{pmatrix} 1 & 1 \\ 1 & -3 \end{pmatrix}$$
with eigenvalues $r_1 = -1+\sqrt{5}$, $r_2 = -1- \sqrt{5}$. The eigenvalues are real, with opposite sign. Hence the critical point of the associated linear system is a saddle, which is unstable.
Attached is the part(d).
After multiplying the second equation by y , it follows that $y=1/-1$ . Hence the critical points of the system are at (-1,1) and (-1,-1).
(b,c) Note that $F(x,y) = 1-xy$ and $G(x,y) = x-y^3$ . The Jacobian matrix of the vector field is $$ J = \begin{pmatrix} -y & -x \\ 1 & -3y^2 \end{pmatrix} $$
At the critical point (1,1), the coefficient matrix of the linearized system is$$ J(1,1) = \begin{pmatrix} -1 & -1 \\ 1 & -3 \end{pmatrix}$$
with eigenvalues $r_1 = r_2 = -2$ . The eigenvalues are real and equal. It is easy to show that there is only one linearly independent eigenvector. Hence the critical point is a stable improper node.
At the point (-1,-1), the coefficient matrix of the linearized system is $$ J(-1,-1) = \begin{pmatrix} 1 & 1 \\ 1 & -3 \end{pmatrix}$$
with eigenvalues $r_1 = -1+\sqrt{5}$, $r_2 = -1- \sqrt{5}$. The eigenvalues are real, with opposite sign. Hence the critical point of the associated linear system is a saddle, which is unstable.
Attached is the part(d).
8
Quiz-7 / Re: Q7-T0301
« on: March 30, 2018, 12:45:51 PM »
(a)The critical points are solns of these questions.
$$y+x(1-x^2-y^2) = 0$$ $$-x+y(1-x^2-y^2)=0$$
Multiply the first equation by y and the second equation by x . The difference of the two equations gives $x^2+y^2=0$. Hence the only critical point is at the origin.
(b,c) With $F(x,y) = y+x(1-x^2-y^2)$ and $G(x,y) = -x+y(1-x^2-y^2)$, the Jacobian matrix of the vector field is
$$ J = \begin{pmatrix} 1-3x^2-y^2 & 1-2xy \\ -1-2xy & 1-x^2-3y^2 \end{pmatrix} $$
At the origin, the coefficient matrix of the linearized system is
$$ J(0,0) = \begin{pmatrix} 1 & 1 \\ -1 & 1\end{pmatrix} $$
with complex conjugate eigenvalues $r_{1} = 1+i$, $r_{2} = 1-i$, . Hence the origin is an unstable spiral.
Attached is the part (d)
$$y+x(1-x^2-y^2) = 0$$ $$-x+y(1-x^2-y^2)=0$$
Multiply the first equation by y and the second equation by x . The difference of the two equations gives $x^2+y^2=0$. Hence the only critical point is at the origin.
(b,c) With $F(x,y) = y+x(1-x^2-y^2)$ and $G(x,y) = -x+y(1-x^2-y^2)$, the Jacobian matrix of the vector field is
$$ J = \begin{pmatrix} 1-3x^2-y^2 & 1-2xy \\ -1-2xy & 1-x^2-3y^2 \end{pmatrix} $$
At the origin, the coefficient matrix of the linearized system is
$$ J(0,0) = \begin{pmatrix} 1 & 1 \\ -1 & 1\end{pmatrix} $$
with complex conjugate eigenvalues $r_{1} = 1+i$, $r_{2} = 1-i$, . Hence the origin is an unstable spiral.
Attached is the part (d)
9
Quiz-6 / Re: Q6--T0801
« on: March 17, 2018, 12:32:00 PM »
Setting $x = \xi t^r$ results in the algebraic equations
$$\begin{pmatrix}1-r & -1\\\ 5 & -3-r\end{pmatrix} \begin{pmatrix}\xi_1 \\ \xi_2 \end{pmatrix} = \begin{pmatrix}0 \\ 0 \end{pmatrix}$$
The characteristic equation is $r^2+2r+2=0$, with root $r = -1+i, r = -1-i$. Substituting reduces $r = -1-i$ the system of equations to $(2+i)\xi_{1}-\xi_{2}=0$,The eigenvectors are $\xi^{(1)} = (1,2+i)^T$, $\xi^{(2)} = (1,2-i)^T$.Hence one of the complex-valued solutions is given by $$x^{(1)} = \begin{pmatrix}1 \\\ 2+i \end{pmatrix}e^{-(1+i)t} =\begin{pmatrix}1 \\\ 2+i \end{pmatrix}e^{-t}(\cos t - i \sin t) $$
Therefore the general solution is $$x = c_1 e^{-t}\begin{pmatrix}\cos t \\\ 2\cos t + \sin t \end{pmatrix} + c_2 e^{-t}\begin{pmatrix}\sin t \\\ -\cos t + 2\sin t \end{pmatrix}$$.
Attached is the graph.
$$\begin{pmatrix}1-r & -1\\\ 5 & -3-r\end{pmatrix} \begin{pmatrix}\xi_1 \\ \xi_2 \end{pmatrix} = \begin{pmatrix}0 \\ 0 \end{pmatrix}$$
The characteristic equation is $r^2+2r+2=0$, with root $r = -1+i, r = -1-i$. Substituting reduces $r = -1-i$ the system of equations to $(2+i)\xi_{1}-\xi_{2}=0$,The eigenvectors are $\xi^{(1)} = (1,2+i)^T$, $\xi^{(2)} = (1,2-i)^T$.Hence one of the complex-valued solutions is given by $$x^{(1)} = \begin{pmatrix}1 \\\ 2+i \end{pmatrix}e^{-(1+i)t} =\begin{pmatrix}1 \\\ 2+i \end{pmatrix}e^{-t}(\cos t - i \sin t) $$
Therefore the general solution is $$x = c_1 e^{-t}\begin{pmatrix}\cos t \\\ 2\cos t + \sin t \end{pmatrix} + c_2 e^{-t}\begin{pmatrix}\sin t \\\ -\cos t + 2\sin t \end{pmatrix}$$.
Attached is the graph.
10
Quiz-5 / Re: Q5--T0501, T5101
« on: March 09, 2018, 06:02:12 PM »
Also, we can Sketch the graph.
(c) Attached is the graph
Edit: Glad to see Junya add the graph afterwards.
(c) Attached is the graph
Edit: Glad to see Junya add the graph afterwards.
11
Quiz-5 / Re: Q5--T0401, T0901
« on: March 09, 2018, 05:57:49 PM »
Solving the first equation for B , we have $x_{2} = x_{1}'/2$. Substitution into the second equation results in $$x_{1}''/2 = -2x_{1}$$
The resulting equation is $x_1'+4x_1 = 0$, with general soln. $$x_{1}(t) = c_{1}cos2t+c_{2}sin2t$$
With $x_2$ given in terms of x_{1}, it follows that $$x_2(t) = -c_{1}sin2t+c_{2}cos2t$$.
Imposing the specific conditions, we have $c_1=3,c_2 = 4$.
Therefore, $x_1(t) = 3cos2t+4sin2t$ and $x_2(t)=-3sin2t+4cos2t$
Attached is the graph.
The resulting equation is $x_1'+4x_1 = 0$, with general soln. $$x_{1}(t) = c_{1}cos2t+c_{2}sin2t$$
With $x_2$ given in terms of x_{1}, it follows that $$x_2(t) = -c_{1}sin2t+c_{2}cos2t$$.
Imposing the specific conditions, we have $c_1=3,c_2 = 4$.
Therefore, $x_1(t) = 3cos2t+4sin2t$ and $x_2(t)=-3sin2t+4cos2t$
Attached is the graph.
12
Quiz-5 / Re: Q5--T0101, T0801
« on: March 09, 2018, 05:53:39 PM »
Solving the first equation for B , we obtain $x_2 = x_1'/2+x_1/4$. Substitution into the second equation results in $$x_1''/2+x_1'/4 = -2x_1-(x_1'/2+x_1'/4)/2$$.
Rearranging the terms, the single differential equation for $x_1$ is $$x_1''+x_1'+\frac{17}{4}x_1=0$$
The general soln is $$x_1(t) = e^{-t/2}(c_1cos2t+c_2sin2t)$$. With $x_2$ given in terms of $x_1$, it has
$$x_2(t) = e^{-t/2}(-c_1cos2t+c_2sin2t)$$.
Imposing the specified initial conditions, we obtain $c_1 = -2, c_2 = 2$. Hence,
$$x_1(t) = e^{-t/2}(-2cos2t+2sin2t)$$.
$$x_2(t) = e^{-t/2}(2cos2t+2sin2t)$$.
Attached is the graph.
Rearranging the terms, the single differential equation for $x_1$ is $$x_1''+x_1'+\frac{17}{4}x_1=0$$
The general soln is $$x_1(t) = e^{-t/2}(c_1cos2t+c_2sin2t)$$. With $x_2$ given in terms of $x_1$, it has
$$x_2(t) = e^{-t/2}(-c_1cos2t+c_2sin2t)$$.
Imposing the specified initial conditions, we obtain $c_1 = -2, c_2 = 2$. Hence,
$$x_1(t) = e^{-t/2}(-2cos2t+2sin2t)$$.
$$x_2(t) = e^{-t/2}(2cos2t+2sin2t)$$.
Attached is the graph.
13
Quiz-3 / Re: Q3-T0801
« on: February 10, 2018, 06:24:35 PM »
The characteristic equation is $$2r^2+r-4r=0$$ with roots $$r = -\frac{1}{4}+\frac{\sqrt{33}}{4}, -\frac{1}{4}-\frac{\sqrt{33}}{4}$$
Therefore, the general equation is $$y = c_{1}exp(-1-\sqrt{33})t/4+c_{2}exp(-1+\sqrt{33})t/4$$
$$y = (-\frac{1}{4}+\frac{\sqrt{33}}{4})c_{1}exp(-1-\sqrt{33})t/4+ (-\frac{1}{4}-\frac{\sqrt{33}}{4})c_{2}exp(-1+\sqrt{33})t/4$$
In order to satisfy the initial conditions, we require that $$c_{1}+c_{2}=0$$
$$(-\frac{1}{4}+\frac{\sqrt{33}}{4})c_{1}+(-\frac{1}{4}-\frac{\sqrt{33}}{4})c_{2}=1$$.
Therefore, we can get $$c_{1}=\frac{-2}{\sqrt{33}}, c_{2}=\frac{2}{\sqrt{33}}$$.
Therefore, the final solution is $$y = \frac{-2}{\sqrt{33}}exp(-1-\sqrt{33})t/4+\frac{2}{\sqrt{33}}exp(-1+\sqrt{33})t/4$$,
y->\infty as t -> \infty
Therefore, the general equation is $$y = c_{1}exp(-1-\sqrt{33})t/4+c_{2}exp(-1+\sqrt{33})t/4$$
$$y = (-\frac{1}{4}+\frac{\sqrt{33}}{4})c_{1}exp(-1-\sqrt{33})t/4+ (-\frac{1}{4}-\frac{\sqrt{33}}{4})c_{2}exp(-1+\sqrt{33})t/4$$
In order to satisfy the initial conditions, we require that $$c_{1}+c_{2}=0$$
$$(-\frac{1}{4}+\frac{\sqrt{33}}{4})c_{1}+(-\frac{1}{4}-\frac{\sqrt{33}}{4})c_{2}=1$$.
Therefore, we can get $$c_{1}=\frac{-2}{\sqrt{33}}, c_{2}=\frac{2}{\sqrt{33}}$$.
Therefore, the final solution is $$y = \frac{-2}{\sqrt{33}}exp(-1-\sqrt{33})t/4+\frac{2}{\sqrt{33}}exp(-1+\sqrt{33})t/4$$,
y->\infty as t -> \infty
14
Quiz-3 / Re: Q3-T0401
« on: February 10, 2018, 06:08:19 PM »
Substitution of the assumed solution $$y = e^{rt}$$ results in the characteristic equation $r^2+4r+3=0$, The roots of the equation are r = -1/-3. Hence the general solution is $$y = c_{1}e^{-t}+c_{2}e^{-3t}$$. Its derivative is $$y' = -c_{1}e^{-t}+-3*c_{2}e^{-3t}$$
Based on the first equation y(0)=1, we can know that $$c_{1}+c_{2} = 2$$.
Based on the second equation y'(0)=-1, we can know that $$-c_{1}-3c_{2}=-1$$, therefore, $$c_{1}= \frac{5}{2}, c_{2} = -\frac{1}{2}$$
Therefore we can get the equation that $$y = \frac{5}{2}e^{-t}-\frac{1}{2}e^{-3t}$$.
y ->0 when t -> \infty
Based on the first equation y(0)=1, we can know that $$c_{1}+c_{2} = 2$$.
Based on the second equation y'(0)=-1, we can know that $$-c_{1}-3c_{2}=-1$$, therefore, $$c_{1}= \frac{5}{2}, c_{2} = -\frac{1}{2}$$
Therefore we can get the equation that $$y = \frac{5}{2}e^{-t}-\frac{1}{2}e^{-3t}$$.
y ->0 when t -> \infty
15
Quiz-3 / Re: Q3-T5101
« on: February 10, 2018, 06:04:13 PM »
The characteristic equation is $r^2-2r-1=0$, with root of $r = 1 + \sqrt{3}, 1- \sqrt{3}$.
Hence the general solution is $$y = c_{1}exp(1-\sqrt{3})t+c_{2}exp(1+\sqrt{3})t$$
Hence the general solution is $$y = c_{1}exp(1-\sqrt{3})t+c_{2}exp(1+\sqrt{3})t$$
Pages: [1] 2