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16

Quiz-5 / Re: Q5--T0301, T0701

« on: March 09, 2018, 06:01:19 PM »

Guanlin's solution is correct! But here's the typed solution
a) Isolate $x_2$ in equation 1 we get
$$x_2 = \frac{3}{2}x_1 - \frac{1}{2}x_1'$$
Differentiate both sides with respect to $t$ we get
$$x_2' = \frac{3}{2}x_1' - \frac{1}{2}x_1''$$
Substitute into the second equation we get $$ \frac{3}{2}x_1' - \frac{1}{2}x_1'' = 2x_1 - 2(\frac{3}{2}x_1 - \frac{1}{2}x_1') $$
Simplify the expression we get $ x_1'' - x_1' -2x_1 = 0$,which is a second order ODE of $x_1$.


b)
Characteristic equation is $r^2 - r - 2 = (r-2)(r+1) = 0$ with roots $r_1 = 2, r_2 = -1$
General solution for $x_1$ is $x_1 = c_1 e^{2t} + c_2 e^{-t}$
Plug in to $x_2 = \frac{3}{2}x_1 - \frac{1}{2}x_1'$ get
$$x_2 = \frac{3}{2} ( c_1 e^{2t} + c_2 e^{-t}) - \frac{1}{2} (2c_1 e^{2t} - c_2 e^{-t}) = \frac{1}{2}c_1e^{2t} + 2c_2e^{-t}$$
So, $$x_1 = c_1 e^{2t} + c_2 e^{-t}$$ $$x_2 = \frac{1}{2}c_1e^{2t} + 2c_2e^{-t}$$
Plug in $x_1(0)=3, x_2(0) = \frac{1}{2} $ to get $$3 = c_1 + c_2 $$  $$\frac{1}{2} = \frac{1}{2}c_1 + 2c_2 $$
Solve the linear system we have
$$c_1 = \frac{11}{3}, c_2 = -\frac{2}{3}$$
That is, $$x_1 = \frac{11}{3} e^{2t} -\frac{2}{3} e^{-t}$$ $$x_2 = \frac{11}{6} e^{2t} -\frac{4}{3} e^{-t}$$

c) see attached graph
Note that as $t \to \infty$, the graph is asymptotic to the line $x_2 = \frac{1}{2} x_1$ in the first quadrant.

17

Quiz-5 / Re: Q5--T0201, T0601

« on: March 09, 2018, 05:59:39 PM »

a) Isolate $x_2$ in equation 1 we get
$$x_2 = \frac{4}{3}x_1' - \frac{5}{3}x_1$$
Differentiate both sides with respect to $t$ we get
$$x_2' = \frac{4}{3}x_1'' - \frac{5}{3}x_1'$$
Substitute into the second equation and simplify, we get $$ x_1'' - \frac{5}{2}x_1' + x_1 = 0 $$
which is a second order ODE of $x_1$.


b)
Characteristic equation is $r^2 - \frac{5}{2} r + 1 = (r - \frac{1}{2})(r - 2) = 0$ with roots $r_1 = \frac{1}{2}, r_2 = 2$
General solution for $x_1$ is $x_1 = c_1 e^{\frac{1}{2}t} + c_2 e^{2t}$
Plug in to $x_2 = \frac{4}{3}x_1' - \frac{5}{3}x_1$ get
$$x_2 = -c_1 e^{\frac{1}{2}t} + c_2 e^{2t}$$
So, $$x_1 = c_1 e^{\frac{1}{2}t} + c_2 e^{2t}$$ $$x_2 = -c_1 e^{\frac{1}{2}t} + c_2 e^{2t}$$
Plug in $x_1(0)=-2, x_2(0) = 1$ to get $$-2 = c_1 + c_2 $$  $$1= -c_1 + c_2 $$
Solve the linear system we have
$$c_1 = -\frac{3}{2}, c_2 = -\frac{1}{2}$$
That is, $$x_1 = -\frac{3}{2} e^{\frac{1}{2}t} -\frac{1}{2} e^{2t}$$ $$x_2 = \frac{3}{2} e^{\frac{1}{2}t} -\frac{1}{2} e^{2t}$$

c) See attached picture
Note that as $t \to \infty$, the graph is asymptotic to the line $x_2 = x_1$ in the third quadrant.

18

Quiz-5 / Re: Q5--T0501, T5101

« on: March 09, 2018, 05:59:13 PM »

a) Isolate $x_2$ in equation 1 we get
$$x_2 = \frac{1}{2}x_1 - \frac{1}{2}x_1'$$
Differentiate both sides with respect to $t$ we get
$$x_2' = \frac{1}{2}x_1' - \frac{1}{2}x_1''$$
Substitute into the second equation and simplify, we get $$ x_1'' + 3 x_1' + 2x_1 = 0 $$
which is a second order ODE of $x_1$.


b)
Characteristic equation is $r^2 +3 r + 2 = (r + 2)(r + 1) = 0$ with roots $r_1 = -2, r_2 = -1$
General solution for $x_1$ is $x_1 = c_1 e^{-2t} + c_2 e^{-t}$
Plug in to $x_2 = \frac{1}{2}x_1 - \frac{1}{2}x_1'$ get
$$x_2 = \frac{3}{2}c_1 e^{-2t} + c_2 e^{-t}$$
So, $$x_1 = c_1 e^{-2t} + c_2 e^{-t}$$ $$x_2 = \frac{3}{2}c_1 e^{-2t} + c_2 e^{-t}$$
Plug in $x_1(0)=-1, x_2(0) = 2$ to get $$-1 = c_1 + c_2 $$  $$2= \frac{3}{2}c_1 + c_2 $$
Solve the linear system we have
$$c_1 = 6, c_2 = -7$$
That is, $$x_1 = 6 e^{-2t} -7 e^{-t}$$ $$x_2 = 9 e^{-2t}  -7 e^{-t}$$

c) See attached image
Note that as $t\to \infty$, the graph approaches the origin in the third quadrant tangent to the line $x_1 = x_2$.

19

Quiz-4 / Re: Q4-T0701

« on: March 02, 2018, 04:31:13 PM »

Solution:
Define $L[y] = t^2y'' - t(t+2)y' + (t+2)y$

Differentiate $y_1(t)=t$ we get $y'_1(t)=1$ and $y''_1(t)=0$.
Plug in to the given DE we see that
$$L[y_1] = t^2\cdot(0) - t(t+2)\cdot(1) + (t+2)\cdot(t) = -t^2 -2t+t^2+2t = 0$$
This shows that $y_1$ is a solution to the corresponding homogeneous equation.

Differentiate $y_2(t)=te^t$ we get $y'_2(t)=(1+t)e^t$ and $y''_2(t)=(2+t)e^t$.
Plug in to the given DE we see that
$$L[y_2] = t^2\cdot(2+t)e^t - t(t+2)\cdot(1+t)e^t + (t+2)\cdot te^t$$ $$= 2t^2e^t + t^3 e^t - (t^3 + 3t^2 + 2t)e^t + t^2 e^t + 2t e^t = 0$$
This shows that $y_2$ is a solution to the corresponding homogeneous equation.

First, let's compute the $W(y_1,y_2)$:
$$W(y_1,y_2) = det \begin{bmatrix}t&te^t\\1& (1+t)e^t\end{bmatrix} = t^2e^t$$

Now consider $Y(t) = u_1t + u_2te^t$ where $u_1$ and $u_2$ are functions of $t$.
Assume $u'_1t + u'_2te^t = 0$ and plug in $Y'$ and $Y''$ into the give DE we can get the following linear system:
$$\begin{bmatrix}t&te^t\\1& (1+t)e^t\end{bmatrix} \begin{bmatrix}u'_1\\u'_2\end  {bmatrix} = \begin{bmatrix}0\\\frac{2t^3}{t^2}\end  {bmatrix}$$
Then,
$$\begin{bmatrix}u'_1\\u'_2\end  {bmatrix} = \frac{1}{t^2e^t} \begin{bmatrix}(1+t)e^t&-te^t\\-1&t\end{bmatrix} \begin{bmatrix}0\\2t\end  {bmatrix} = \begin{bmatrix}-2\\2e^{-t}\end{bmatrix}$$

Integrate $u_1$ and $u_2$ we get:
$$u_1 = \int{-2 dt} = -2t$$ $$u_2 = \int{2e^{-t} dt} = -2e^{-t}$$

Then $Y(t) = -2t(t) + (-2e^{-t})(te^t) = -2t^2 -2t$ is a particular solution to the given nonhomogeneous equation.
Notice that $L[-2t] = 0$, thus $Y(t) = -2t^2$ is also a particular solution to the given nonhomogeneous equation.

20

Quiz-4 / Q4-T0701 / T0401

« on: March 02, 2018, 04:30:46 PM »

Verify that the given functions $y_1$ and $y_2$ satisfies the corresponding homogeneous equation; then find a particular solution of the given nonhomogeneous equation.
$$t^2y'' - t(t+2)y' + (t+2)y = 2t^3, t>0; y_1(t)=t, y_2(t)=te^t$$

21

Quiz-2 / Re: Q2-T5101

« on: February 02, 2018, 05:56:29 PM »

First, let's show the given DE $x^{2}y^{3} + x(1+y^{2})y' = 0$ is not exact.
Define $M(x,y)=x^{2}y^{3}$, $N(x,y)=x(1+y^{2})$
$$M_y = \frac{\partial}{\partial y}[x^{2}y^{3}] = 3x^{2}y^{2}$$ $$N_x = \frac{\partial}{\partial x}[x(1+y^{2})] = 1+y^{2}  $$
Since $3x^{2}y^{2} ≠ 1+y^{2}$, this implies the given DE is not exact.



Now, let's show that the given DE multiplied by the integrating factor $\mu(x,y) = \frac{1}{xy^{3}}$ is exact.
That is to show $$\frac{1}{xy^{3}}x^{2}y^{3} + \frac{1}{xy^{3}}x(1+y^{2})y' = x + (y^{-3}+y^{-1})y'= 0$$ is exact.

Define $M'(x,y) = x$, $N'(x,y) = y^{-3}+y^{-1}$
Since
$$M'_y = \frac{\partial}{\partial y}(x) = 0 $$ $$N'_x = \frac{\partial}{\partial x}[y^{-3}+y^{-1}] = 0 $$
By theorem in the book, we can conclude that $x + (y^{-3}+y^{-1})y'= 0$ is exact.


Thus, we know there exists a function $\phi(x,y)=C$ which satisfies the given DE.
Also,
$$ \frac{\partial \phi}{\partial x} = x $$ $$ \frac{\partial\phi}{\partial y} = y^{-3}+y^{-1}$$
Integrate $\frac{\partial \phi}{\partial x} = x $ with respect to $x$ we have
$$\phi(x,y) = \frac{1}{2}x^{2} + g(y)$$
Take derivative on both sides with respect to $y$ we get
$$\frac{\partial\phi}{\partial y} = g'(y)$$
Since we know that $\frac{\partial\phi}{\partial y} =  y^{-3}+y^{-1}$
Then $g'(y) = y^{-3}+y^{-1}$
Integrate with respect to $y$ we have
$g(y) = -\frac{1}{2}y^{-2} + ln|y| + C$
Altogether, we have $\phi(x,y) = \frac{1}{2}x^{2} -\frac{1}{2}y^{-2} + ln|y| = C $, which means
$$\frac{1}{2}x^{2} -\frac{1}{2}y^{-2} + ln|y| = C$$
is the general solution to the given DE.

Besides, notice that the constant function $y(x)=0$ $\forall x$ is also a solution to the given DE.

22

Quiz-2 / Re: Q2-T0801

« on: February 02, 2018, 05:02:15 PM »

Since the given DE is not exact, we need to find a function $\mu (x,y)$ such that the equation $\mu (3x^{2}y+2xy+y^{3}) + \mu (x^{2}+y^{2})y’=0$ is exact.

This means that
$$\frac{∂}{∂y}[\mu (3x^{2}y+2xy+y^{3})]=\frac{∂}{∂x}[\mu (x^{2}+y^{2})]$$
$$\frac{∂\mu}{∂y}(3x^{2}y+2xy+y^{3})+\mu (3x^{2}+2x+3y^{2}) = \frac{∂\mu}{∂x}(x^{2}+y^{2}) + \mu (2x)$$
$$\frac{∂\mu}{∂y}(3x^{2}y+2xy+y^{3})+\mu (3x^{2}+3y^{2}) = \frac{∂\mu}{∂x}(x^{2}+y^{2})$$

Suppose $\mu$ is a function of $x$ only. (i.e. $\mu=\mu (x)$)
Then $\frac{∂\mu}{∂y}=0$ and $\frac{∂\mu}{∂x}=\frac{d\mu}{dx}$

Then $$\mu (3x^{2}+3y^{2}) = \frac{d\mu}{dx}(x^{2}+y^{2})$$

Divide both sides by $(x^{2}+y^{2})$ we have $$3\mu = \frac{d\mu}{dx}$$ which is a separable equation.

$$\int{\frac{1}{\mu}d\mu} = \int{3 dx}$$
$$ln(\mu)=3x$$
$$\mu = e^{3x}$$

Thus, $\mu = e^{3x}$ is an integration factor for the given DE.

Multiply the given DE by $\mu = e^{3x}$, we get
$$e^{3x} (3x^{2}y+2xy+y^{3}) + e^{3x} (x^{2}+y^{2})y’=0$$ which is exact.

Then by theorem, we know there exist a function $\phi(x,y)=C$ which satisfies the given DE

We also know that
$\frac{∂\phi}{∂x} = e^{3x} (3x^{2}y+2xy+y^{3})$ and $\frac{∂\phi}{∂y} =e^{3x} (x^{2}+y^{2})$

Integrate $\frac{∂\phi}{∂y} =e^{3x} (x^{2}+y^{2})$ with respect to $y$ we have
$$\phi(x,y)= e^{3x} (x^{2}y+\frac{1}{3}y^{3}) +g(x)$$
Then, take derivative on both sides with respect to $x$, we have
$$\frac{∂\phi}{∂x} = 3e^{3x} (x^{2}y+\frac{1}{3}y^{3}) + e^{3x} (2xy) + g’(x)$$
Simplify a little bit we have
$$\frac{∂\phi}{∂x} = e^{3x} (3x^{2}y + y^{3} + 2xy) + g’(x)$$

Since we know that
$\frac{∂\phi}{∂x} = e^{3x} (3x^{2}y+2xy+y^{3})$

This means that $g’(x)=0$. So $g(x)=C$

Altogether, we have $$\phi(x,y)= e^{3x} (x^{2}y+\frac{1}{3}y^{3})=C$$
Which means the general solution to the given DE is $$e^{3x} (x^{2}y+\frac{1}{3}y^{3})=C$$

23

Quiz-1 / Q1-T0701

« on: January 26, 2018, 02:02:06 PM »

Question: Find the solution of the given initial value problem.
$$y'-2y=e^{2t}, y(0)=2$$
Solution:
Notice that the given DE is a first order linear ODE.
Let $\mu(t)$ denote an integrating factor for the given DE.
$$\mu(t) = e^{\int -2 dt} = e^{-2t} $$
Multiply the given DE by $\mu(t)$: note that $\mu(t) ≠ 0$ for all $t$
$$e^{-2t}y'- 2e^{-2t}y = e^{-2t} e^{2t}$$
Simplify the equation:
$$\frac{d}{dt} (e^{-2t}y)= 1$$
Integrate both sides with respect to $t$:
$$e^{-2t}y = t + C$$
Isolate $y$:
$$y = (t+C)e^{2t}$$
Since $y(0)=2$, then $2 = (0+C)\cdot e^{0} = C\cdot 1 = C$
Thus, solution to the given IVP is
$$y = (t+2)e^{2t}$$

24

Quiz-1 / Re: Q1-T5101

« on: January 25, 2018, 09:41:08 AM »

First notice that the given DE is a first order linear differential equation.
Rewrite it in the standard form we have
$$y'- \frac{1}{t}y = te^{-t}$$

Let $\mu(t)$ denote the integrating factor.
$$\mu(t)=e^{\int -\frac{1}{t} dt} = e^{- log(t)} = e^{log((t)^{-1})} = t^{-1}. $$

Multiply both sides of the equation by $\mu(t)$ we get
$$t^{-1}y'-t^{-2}y=e^{-t}$$
$$\frac{d}{dx}[t^{-1}y]=e^{-t}$$

Integrate both sides with respect to $t$ we get
$$t^{-1}y = -e^{-t} + C$$

So,
 $$y=-te^{-t} + Ct$$

When $t \to \infty$:
Case1: if $C=0$
Then by one use of L'Hopital's rule, we get $y \to 0$.

Case2: if $C>0$
Then $y \to +\infty$.

Case3: if $C<0$
Then $y \to - \infty$.

25

Quiz-1 / Re: Q1-T0201

« on: January 25, 2018, 09:13:56 AM »

26

Quiz-1 / Re: Q1-T0801

« on: January 25, 2018, 08:58:51 AM »