#33.

OK I assumea u subst so we can separate

$$\dfrac{dy}{dx}= \dfrac{y/x-3}{2-y/x} $$]]>

ok I did this first

$u(t)y'+u(t)y=u(i)5\sin{2t}$

then

$\frac{1}{5}u(t)y'+\frac{1}{5}u(t)y=u(i)\sin{2t}$

so far .... couldn't find an esample to follow

____________________________________________

book answer

$y=ce^{−t}+\sin 2t−2\cos 2t$

$\textit{y is asymptotic to sin2t−2cos2t as $t\to\infty$}$]]>

$\dot{X_1}=2\cos X_2, X_1(0)=a$

$\dot{X_2}=3\sin X_1, X_2(0)=b$

has a unique solution for the arbitrary constants $a$ and $b$.

how to solve this system? Thanks.]]>

$\check{X_1}=X_1$

$\check{X_2}=aX_2$

where $a$ is a constant.]]>

Pleas me a idea]]>

ok not sure if this is the best first steip,,,, if so then do a $u=\dfrac{x}{y}$ ?]]>

OK going to do #31 if others new OPs

I went over the examples but???

well we can't 6seem to start by a simple separation

I think direction fields can be derived with desmos]]>

well each one is a little different so,,,

$$\dfrac{dy}{dt}=\dfrac{ty(4-y)}{3},\qquad y(0) =y_0$$

not sure if this is what they meant on the given expression]]>

OK going to comtinue with these till I have more confidence with it

$$\dfrac{dy}{dx}=2 (1+x) (1+y^2), \qquad y(0)=0$$

separate

$$(1+y^2)\, dy=(2+2x)\, dx$$]]>

it's late so I'll just start this

\[ \dfrac{dy}{dx}=\dfrac{2\cos 2x}{3+2y} \]

so \[(3+2y) \, dy= (2\cos 2x) \, dx\]

$y^2 + 3 y= sin(2 x) + c$]]>

$2y'+y=3t^2$

Rewrite

$y'+\frac{1}{2}y=\frac{3}{2}t^2$

So

$u(t) = exp\left(\int \frac{1}{2} dt \right)=e^{t/2} $

Hopefully so far]]>

I'm unsure how to begin and solve this question. Any help would be appreciated, thanks.]]>

a) 4

b) 3

c) 5

d) Maximum value does not exist.

I am quite lost on this one. After some thought I am convinced that the maximum value should exist, though I do not have a good argument to support this claim.]]>

Problem: Find a (limited?) solution to the diff eq.

At the end of the solution, when you transform \(\displaystyle \frac{-1}{s+1} + \frac{2}{s-3}\)

why doesnt it become \(\displaystyle -e^{-t} + 2e^{3t} \), t>0 ?]]>

Find the general solution to the system of differential equations

\begin{align*}\displaystyle

y'_1&=y_1+5y_2\\

y'_2&=-2y_1+-y_2

\end{align*}

why is there a $+-y_2$ in the given

ok going to take this a step at a time... so..

$A=\left[\begin{array}{c}1 & 5 \\ -2 & -1 \end{array}\right]$

then

$\left[\begin{array}{c}1-\lambda & 5 \\ -2 & -1-\lambda \end{array}\right]

=\lambda^2+9$ ????]]>

rewrite as

$y^3-3y^2+2y=0$

then factor

$y(y-2)(y-1)=0$

so then

$r=0,1,2$

OK this one is probably obvious but I was wondering why y isn't referred to r in the first place?]]>

Use rescaling to solve: $\phi(x,\epsilon) = \epsilon x^2 + x+1 = 0$ and $\varphi (x,\epsilon) = \epsilon x^3+ x^2 - 4=0$.

I'll write my attempt at solving these two equations, first the first polynomial.

According to the book I first need to find an undetermined gauge that solves the above equation when $\epsilon = 0$, which means I need to guess $x \approx -1 +y \delta_1(\epsilon)$, now I plug it...

Rescaled Coordinates in a polynomial equation.]]>

$$y''+5y'+6y=e^x$$

into a system of first order (nonhomogeneous) differential equations and solve the system.

the characteristic equation is

$$\lambda^2+5\lambda+6=e^x$$

factor

$$(\lambda+2)(\lambda+3)=e^x$$

ok not real sure what to do with this $=e^x$ thing]]>

$-\frac{17}{21}*(3x-2y)+ln(119y-34x-48)=C$. Now how to solve this answer for y(x)?

I also want to know its step by step solution.

If any member knows the correct answer, he/she may reply with correct answer. Your Wolfram Alpha DiffEq calculator gives this answer...

Solving 1st order non-linear ODE]]>

$\begin{cases}

y'_1&=2y_1+y_2-y_3 \\

y'_2&=3y_2+y_3\\

y'_3&=3y_3

\end{cases}$

let

$y(t)=\begin{bmatrix}{y_1(t)\\y_2(t)\\y_3(t)}\end{bmatrix}

,\quad A=\begin{bmatrix}

2 & 1 & -1 \\

0 & 3 & 1 \\

0 & 0 & 3

\end{bmatrix}$

so

\begin{align*}\displaystyle

y_1&=c_1e^{2t}+c_2e^{t}+c_3e^{-t}\\

y_2&=c_2e^{3t}+c_3e^{t}\\

y_3&=c_3e^{3t}

\end{align*}

ok if correct so far...

16.1 Find the general solution to the system of differential equations]]>

$$u''+0.5u'+2u=0$$

ok from examples it looks all we do is get rid of some of the primes and this is done by substitution

so if $u_1=u$ and $u_2=u'_1$

then $u_2=u'$ and $u'_2=u''$

then we have $u'_2+0.5u_2 +2u_1 = 0$

then isolate $u'_2$ thus $u'_2=-0.5u_2-2u_1$

ok the next problem is

$$u''+0.5u'+2u=3\sin t$$ so ???]]>

$\quad y''+5y'+6y=0$

(a)convert into a system of first order (homogeneous) differential equation

(b)solve the system.

ok just look at an example the first step would be

$\quad u=y'$

then

$\quad u'+5u+6=0$

so far perhaps???]]>

$A^{-1}=\begin{bmatrix}1&3\\2&5 \end{bmatrix}$

Solve the matrix equation $AX=B$ to find $x$ and $y$ where

$X=\begin{bmatrix}x\\y \end{bmatrix}\& \quad B=\begin{bmatrix}1\\3 \end{bmatrix}$

ok well first find A

$A=\begin{bmatrix}1&3\\2&5 \end{bmatrix}^{-1}

=\left[ \begin{array}{rr|rr}1&3&1&0 \\ 2&5&0&1\end{array}\right]

=\left[ \begin{array}{rr|rr}1&0&-5&3 \\ 0&1&2&-1\end{array}\right]

=\left[ \begin{array}{rr} -5 & 3 \\ 2 & -1 \end{array} \right]$

then we...

5.5 Solve the matrix equation AX=B to find x and y]]>

\begin{align*}\displaystyle

y'_1&=2y_1+3y_2+5x\\

y'_2&=y_1+4y_2+10

\end{align*}

rewrite as

$$Y'=\left[\begin{array}{c}2 & 3 \\ 1 & 4 \end{array}\right]Y

+\left[\begin{array}{c}5x\\ 10\end{array}\right]$$

ok not sure what to do with this]]>

Solving Rational ODE's of the form (ax+by+c) dx+(ex+fy+g) dy=0.]]>

$$\displaystyle y^{\prime\prime} + 5y^\prime + 6y =0$$

ok I presume this means to find a general solution so

$$\lambda^2+5\lambda+6=(\lambda+3)(\lambda+2)=0$$

then the roots are

$$-3,-2$$

thus solutions

$$e^{-3x},e^{-2x}$$

ok I think the Wronskain matrix is next which I have never done.

well so far,,,, any suggest...

also this might be DE question ... not sure..]]>

$6y''-5y'+y=0\quad y(0)=4 \quad y'(0)=0$

if $r=e^{5t}$ then

$\displaystyle 6y''-5y'+y=(r-3)(r-2)=0$

then

$y=c_1e^{3t}+c_1e^{2t}=0$

for $y(0)=4$

$y(0)=c_1e^{3(0)}+c_1e^{2(0)}=4$

ok I don't see how the last few steps lead to the book answer (11)

]]>

Given #11

$\quad\displaystyle

xdx+ye^{-x}dy=0,\quad y(0)=1$

a. Initial value problem in explicit form.

$\quad xdx=-ye^{-x}dy$

separate

$\quad \frac{x}{e^{-x}}\, dx=-y\, dy$

simplify

$\quad xe^x\, dx=-y\, dy$

rewrite

$\quad y\,dy=-xe^x\,dx$

integrate (with boundaries)

$\quad \int_1^y u\,dy=-\int_0^x ve^v\,dv$

OK i did this so far hopefully ok but didnt know how to do b and c (on desmos)

b. Plot the graph of the solution

c. Interval of solution.]]>

so far

I could not get to the W|A solution before applying the y(0)=a

here is the book answer for the rest

]]>

Solve the initial value problem

$Y'=\left|\begin{array}{rr}2 & 1 \\-1 & 2 \end{array}\right|Y

+\left|\begin{array}{rr}e^x \\0 \end{array}\right|,

\quad Y(0)=\left|\begin{array}{rr} 1 \\1 \end{array}\right| $

ok so we have the form $y'=AY+G$

rewrite as

$$\displaystyle

\left|\begin{array}{rr}y_1^\prime \\y_2^\prime \end{array}\right|

=\left|\begin{array}{rr}2 & 1 \\-1 & 2 \end{array}\right|

\left|\begin{array}{rr}y_1 \\y_2\end{array}\right|

+\left|\begin{array}{rr}e^x \\0...

31.6 Solve the initial value problem]]>

cordially]]>

where $x,y,z$ are all functions of the variable t

\begin{align*}\displaystyle

x'&=-4x-3y+3z\\

y'&=3x+2y-3z+e^t\\

z´&=-3x-3y+2z

\end{align*}

write a system in the matrix form $Y'=AY+G$]]>

$x'=x+\sin(t)$

ok this uses x rather than y which threw me off

so rewrite as

$x'-x=\sin(t)$

thus $u(t)=e^{-t}$

$e^{-t}x'-e^{-t}x=e^{-t}\sin{t}$

and

$(e^{-t}x)'=e^{-t}\sin{t}$

intergrate thru

$\displaystyle e^{-t}x=\int {e^{-t} \sin(t)} dt = -1/2 e^{-t}\sin(t) - 1/2 e^{-t} \cos(t) +e^{-t} c$

divide thru

$\displaystyle x(t)=-\frac{\sin t}{2}-\frac{\sin t}{2}+ce^t$

ok well typos probably

W|A returned $\displaystyle x(t) = c_1 e^t - \frac{\sin(t)}{2} -...

7.1 x'=x+sin(t) solve de]]>

use the different techniques of diagonal, diagonalizedable, or triangular.

$\begin{cases}

y'_1 & =3y_1 \\ y'_2 & =2y_2\end{cases}$

set matrix

$A= \begin{pmatrix}0 &3\\0 &2\end{pmatrix}$

then find eigenvalues

ok just seeing if i am starting out correctly with this..

not sure what the difference is between diagonal and diagonalizedable? but it looks diagonal]]>

$$\frac{1}{G_{zx}}\frac{\partial ^2\phi (x,y)}{\partial^2 y}+\frac{1}{G_{zy}}\frac{\partial ^2\phi (x,y)}{\partial^2 x}=-2 \theta$$

where $G_{zy}$, $G_{zx}$, $\theta$, $a$, and $b$ are constants and with BCs $\phi (0,y)=\phi (a,y)=0$ and $\phi (x,-b)=\phi (x,b)=0$. The solution that I have sets $\phi (x,y)=\sum _{k=1,3,5\text{...}}^{\infty } Y(y) \sin \left(\frac{\pi...

How to get a converging solution to a second-order PDE?]]>

$$Y'=\begin{bmatrix}2 & 1 & 0 \\0 & 2 & 1 \\ 0 & 0 & 4 \end{bmatrix}Y$$

subtract $\lambda$ from the diagonal entries of the given matrix and take det:

$$\left|

\begin{array}{ccc}

- \lambda + 2 & 1 & 0 \\

0 & - \lambda + 2 & 1 \\

0 & 0 & - \lambda + 4

\end{array}\right|

=(-\lambda+2)^{2}(-\lambda+4)$$

the roots are:

$$\lambda_1=2,\quad\lambda_2=2, \quad\lambda_3=4$$

this is the example in the book I am trying to follow but I don't see...

28.2 Solve the system y'={}Y]]>

Please help me convert my equation to this form and tell me how it can be done or at least name...

Partial Differential Equation - solve with Matlab]]>

Determine the equations of the orthogonal trajectories of the following family of curve;

e^{x}(xcosy - ysiny) = c

]]>

in (3) they say any polynomial of degree less than 2, yet the example is degree 2 ???

ok this is due tomro, so hopefully I can get a handle on it today

I think this is familiar, just is the notation is stumping me.

really appreciate any insight...

did a screenshot to avoid typos]]>

I have a question regarding problem 3a on page 9 in Omalley's Singular Perturbation Methods for ODEs, regarding what he called "Friedrich's problem".

I am not sure how did they get the asymptotic relation: $x(t,\epsilon) \sim (\exp(1-t)-\exp(1-t/\epsilon))$ as $\epsilon \to 0$ uniformly in $t \in [0,1]$.

I get that the solution is

\begin{gather}\nonumber x(t,\epsilon)=(1/(\exp((-1+\sqrt{1-4\epsilon})/(2\epsilon))-\exp((-1-\sqrt{1-4\epsilon})/(2\epsilon)))\\ \nonumber...

Friedrich's problem.]]>

Solve the initial value problem and find the critical value $\displaystyle a_0$ exactly.

$\displaystyle 3y'-2y=e^{-\pi/2}\quad y(0)=a$

divide by 3

$\displaystyle y'-\frac{2}{3}y=\frac{e^{-\pi/2}}{3}$

with integrating factor $\displaystyle e^{-2t/3}$ multiply through

$\displaystyle e^{-2t/3}y'-\frac{2}{3} e^{-2t/3}y=\frac{e^{-\pi/2}}{3}e^{-2t/3}$

ok just seeing if this is going in the right direction]]>

Find the general solution of the given differential equation, and use it to determine how

solutions behave as $t\to\infty$.

$2y'+y=3t$

divide by 2

$y'+\frac{1}{2}y=\frac{3}{2}t$

find integrating factor,

$\displaystyle\exp\left(\int \frac{1}{2} dt\right)=e^{t/2}+c$

multiply thru

$e^{t/2}y'+e^{t/2}\frac{y}{2}

=\frac{3e^{t/2}}{2}t $

ok something went

------------------------------------...

2.1.9 Find the general solution of the given differential equation]]>

$ y = t^r$

for

$t > 0 $}$

$t^2y''+4ty'+2y = 0$

$\color{red}{r=-1,-2}$

$t^2y''-4ty'+4y=0$

$\color{red}{r=1,4}$

ok the book answers are in red, and obviously this is from a factored quadradic but I couldn't an example of the steps.

probably just a couple!]]>

Thank you!]]>

The equation is proven to be homogeneous. However after using substitution of y=zx and its' derivative, I was not able to separate the variables conveniently as shown. Please advise. Thank you!]]>

I do not know how to begin with the following problem:

Suppose that $f$ is $2\pi$-periodic and let $a$ be a fixed real number. Define $F(x)=\int_{a}^{x} f(t)dt$, for all $x$ .

Show that $F$ is $2\pi$-periodic if and only if $\int_{0}^{2\pi}f(t)dt=0$.

Thanks,

Cbarker1]]>

I am confused about how to start with the following problem: using the solution from ex. 3:

$u(x,t)=F(x+ct)+G(x-ct)$

"For data u(x,0)=0 and ${u}_{t}=\frac{x}{(x^2+1)^2}$ where x is from neg. infinity to pos. infinity."

Thanks

Cbarker1]]>

I do not know how to begin with the following problem:

you are asked to solve the wave equation subject to the boundary conditions ($u(0,t)=u(L,t)=0$), $u(x,0)=f(x)$ for $0\le x\le L$ and ${u}_{t}(x,0)=g(x)$ for $0\le x\le L$ . Hint: using the $u(x,t)=\sum_{n=1}^{\infty}{{b}_{n}\sin(\frac{n\pi x}{L})\cos(\frac{n\pi c t}{L})}$ and the remark is that if the initial velocity is nonzero, then additional terms of the form ${{{b}^{*}}_{n}\sin(\frac{n\pi x}{L})\cos(\frac{n\pi c...

Wave Equation Question Part 3]]>

I am confused about how to start with the following problem:

Consider an electrical cable running along the x-axis that is not well insulated from the ground, so that leakage occurs along its entire length. Let V(x,t) and I(x,t) denote the voltage and current at point x in the wire at time t. These functions are related to each other by the system

${V}_{x}=-L{I}_{t}-RI$ and ${I}_{x}=-C{V}_{t}-GV$ where L is the inductance, R is the resistance, C is the capacitance, and G...

Telegraph Equation Verification]]>

Hi. I do not how to begin for the following question:

Ex. 5. Using the solution in Ex. 3, solve the wave equation with initial data

$u(x,t)=\frac{1}{{x}^2+1}$ and $\pd{u}{t}(x,0)=0$ for $x\in(-\infty,\infty)$.

The solution, (I have derived this solution in Ex. 4), that is given in Ex. 3 is the following: $u(x,t)=F(x+ct)+G(x-ct)$

Thanks,

Cbarker1]]>

Consider the logistic equation

$$\dot{y}=y(1-y). $$

(a) Find the solution satisfying $y_1(0)=6$ and $y_2(0)=−1. $

$y_1(t)= ?$

$y_2(t)= ?$

(b) Find the time $t$ when $y_1(t)=3. $

$t= ?$

(c) When does $y_2(t)$ become infinite?

$t= ?$]]>

There is an example in my textbook worded as follows;

A particle of mass 2kg moves along the positive x axis under the action of a force directed towards the origin. At time t seconds, the displacement of P from O is x metres and P is moving away from O with a speed of v ms^-1. The force has magnitude 40x N. The particle P is also subject to resistive forces of magnitude 16v N.

a) Show that the equation of motion of P is d^2x/dt^2 + 8 dx/dt +20x = 0

I do not understand how the force...

2nd order differential equation - equation of motion]]>

to find approximate values of the solution at tt = 0.1, 0.2, 0.3, 0.4, and 0.5.

Compare them to the exact values...

Euler’s Method y'+2y=2-e^(-4t) y(0)-1]]>

$y' + 2y = 2 - {{\bf{e}}^{ - 4t}} \hspace{0.25in}y\left( 0 \right) = 1$

find

$\displaystyle u(x)=\exp \left(\int 2 \, dt\right) =e^{2t}$

$e^{2t}y' + 2\frac{e^{2t}}{2}y =2 e^{2t} - {{\bf{e}}^{ - 4t}}e^{2t}$

$(e^{2t}y)'=2e^{2t}-e^{-2t}$

i continued but didn't get the answer which is

$y\left( t \right) = 1 + \frac{1}{2}{{\bf{e}}^{ - 4t}} - \frac{1}{2}{{\bf{e}}^{ - 2t}}$]]>

$$y'=\frac{3t^2}{3y^2-4}, \quad y(1)=0$$

find estimates for t =0.1; 0.2; 0.4

using step sizes of h = 0.1; 0.05; and 0.025

using Euler's Method

ok ran out of time to do more steps but asume separation of variables is next]]>

(a) Find approximate values of the solution of the given initial value problem\\

at $t = 0.1, 0.2, 0.3$, and $0.4$ using the Euler method with $h = 0.1$.

(b) Repeat part (a) with h = 0.05. Compare the results with those found in (a).

(c) Repeat part (a) with h = 0.025. Compare the results with those found in (a) and (b).

(d) Find the solution $y=\phi(t)$ of the given problem and evaluate

$\phi(t)$ at $t = 0.1,\quad 0.2, \quad 0.3,$ and $0.4$.

#1. $\quad\displaystyle...

2.7.2 Euler's method]]>

$\displaystyle \frac{dy}{dt}=ay-b$}$

(a) Find the equilibrium solution $y_e$

rewrite as

$y'-ay=b$

$\displaystyle -\exp\int a \, da=e^{a^{2}/2}$

$\color{red}{y_e=b/a}$

(b) Let $Y(t)=y-y_e$; thus $Y(t)$ is the deviation from the equilibrium solution.

Find the differential equation satisfied by $Y(t)$.

????

$\color{red}{Y' = aY}$

ok stopped in my tracks.. red is book answer]]>

y'&=y-5\quad y(0)=y_0\tag{given}\\

y'-y&=-5\\

u(x)&=\exp\int-1\, dx = e^{-t}\\

(e^{-t}y)&=-5e^{-t}\\

e^{-t}y&=-5\int e^{-t} dt = -5e^{-t}+c\\

&y=-5\frac{e^{-t}}{e^{-t}}+\frac{c}{e^{-t}}\\

y&=\color{red}{5+(y_0-5)e^t}

\end{align*}$

this is similiar to one I posted before but can't seem to get the book answer (red)

the $y_0$ is ???]]>

rewrite

$\quad\displaystyle y'-2y=-10$

obtain u(x)

$\quad\displaystyle u(t)=\exp\int -2 \, dt=e^{-2t}$

integrate using the boundaries:

$\quad\displaystyle \int_{y_0}^{e^{-2t}y}\,du=-10\int_{0}^{t}e^{-2v}\,dv$

resulting in

$\quad\displaystyle e^{-2t}y-y_0=-5\left(e^{-2t}-1\right)$

multiply thru by $e^{2t}$

$\quad\displaystyle y-y_0e^{2t}=5\left(1-e^{2t}\right)$

then

$\quad\displaystyle...

2.1.1c Solve for the following initial value]]>

rewrite

$$y'-2y=-5$$

obtain u(x)

$$u(x)=\exp\int-2\, dx = e^{-2t}$$

then

$$(e^{-2t}y')=5e^{-2t}$$

just reviewing but kinda

]]>

doing #1

ok first I divided thru

$$y' + \frac{\ln{t}}{t-3}y=\frac{2t}{t-3}$$

but the $$\exp\int p(t) \, dt$$ step kinda baloated???

ok I see the denominator has $t-3$ so presume 3 is one of the interval ends

but why 0. ?

book answwer is

]]>

$$\quad\displaystyle

y^2(1-x^2)^{1/2} \,dy = \arcsin{x}\,dx,

\quad y(0) = 1$$

separate

$$y^2 \,dy = \frac{\arcsin{x}}{(1-x^2)^{1/2}}\,dx,$$

Integrate

\begin{align*}

\int y^2 \,dy& = \int\frac{\arcsin{x}}{\sqrt{(1-x^2)}}\,dx,

\end{align*}

ok I assume a trig substitution to solve

book answer

]]>

\(\displaystyle \ddot{ \theta} = A~\cos( \theta )\)

and

\(\displaystyle \dot{ \theta } ^2 = B( \sin( \theta ) - \sin( \theta _0 ) )\)

where the dot represents a time derivative and \(\displaystyle \theta _0\) is the angle at time t = 0. (This is the harmonic oscillator where the angle is not taken to be small. A and B are related constants and I can give the derivations if you feel you need...

Iterating an equation]]>

y^{\prime}=

\frac{e^{-x}-e^x}{3+4y},

\quad y(0)=1$

rewrite

$\frac{dy}{dx}=\frac{e^{-x}-e^x}{3+4y}$

separate

$3+4y \, dy = e^{-x}-e^x \, dx$

integrate

$2y^2+3y=-e^{-x}-e^x+c$

well if so far ok presume complete the square ???

book answer

$(a)\quad y=-\frac{3}{4}+\frac{1}{4}

+\sqrt{65-8e^x-8e^{-x}}$\\

$(c)\quad|x|<2.0794\textit{ approximately}$]]>

$\displaystyle y^{\prime}= \frac{x(x^2+1)}{4y^3}, \quad y(0)=-\frac{1}{\sqrt{2}}$

rewrite

$$\displaystyle\d{y}{x}=\frac{x(x^2+1)}{4y^3}$$

$$\displaystyle y^3\, dy=\frac{x(x^2+1)}{4}\, dx$$

integrate

$$\displaystyle\frac{y^4}{4}= \frac{1}{4}\left(\frac{x^4}{4} +\frac{ x^2}{2}\right)$$

$$\displaystyle y^4=\frac{x^4}{4} +\frac{ x^2}{2}$$

so far hopefully sorta?]]>

from i would deduct that $dr=r^2$ and $d\theta = \theta then$

$$\d{r}{\theta}=\frac{\theta}{r^2}

\text{ or }

\frac{1}{r^2}dr=\frac{1}{\theta}d\theta$$

intregrate]]>

(a) find initial value (b)plot and (c) interval

$$\displaystyle

y^{\prime}=2x/(y+x^2y), \quad y(0)=-2$$

separate the variables

$$\frac{dy}{dx}=\frac{2x}{y+x^2y}=\frac{2x}{y(1-x^2)}$$

$$y\, dy =\frac{2x}{(1-x^2)}dx$$

integrate

$$\int y\, dy = -\sqrt{4}\int \frac{x}{(1-x^2)} \, dx$$

$$\frac{y^2}{2}= -2\frac{\ln(1 - x^2)}{2}$$

$$y=-\sqrt{4}\sqrt{\ln(1-x^2)}$$

ok Im stuck again.,...

book answer

$$(a)\, y = −[2 ln(1 + x^2) +...

2.2.12 (a) find initial value (b)plot and (c) interval]]>

$$xdx+ye^{-x}dy=0, \quad y(0)=1$$

\begin{align*}\displaystyle

xdx&=-ye^{-x}dy \\

\frac{x}{e^{-x}}\, dx&=-y\, dy\\

xe^x\, dx&=-y\, dy

\end{align*}

(b) Plot the

$\quad \textit{ok... I tried some attempts in W|A but my input didn't work}\\$

(c) Determine (at least approximately) the

$\quad \textit{...provided by...

de2.2.11 initial value, graph, intervals]]>

$\vec{x'}=\small\begin{pmatrix}1&2\\3&2\end{pmatrix}\vec{x}+t\small\begin{pmatrix}2\\-4\end{pmatrix}$

Now i got the solution to this differential equation system as

$\vec{x}(t)=c_1e^{-t}\small\begin{pmatrix}-1\\1\end{pmatrix}$+$c_2e^{4t}\small\begin{pmatrix}2\\3\end{pmatrix}$+$t\small\begin{pmatrix}3\\\frac{-5}{2}\end{pmatrix}$+$\small\begin{pmatrix}-2.75\\2.875\end{pmatrix}$

Now i converted this differential equation system into ordinary differential equation...

differences in the solutions of DE system and its converted 2nd order Differential equation]]>

let \(\displaystyle U =X(x)T(t) \)

so

\(\displaystyle 4X\frac{\partial T}{\partial t}+T\frac{\partial X}{\partial x} = 3XT\)

\(\displaystyle 4\frac{\partial T}{T \partial t}+\frac{\partial X}{X \partial x} = 3\)

\(\displaystyle \left( 4\frac{\partial T}{T \partial t}-3 \right) +\frac{\partial X}{X \partial x} = 0 \)

let K = constant

\(\displaystyle \frac{\partial X}{X \partial x} =\left( 3 -...\)

Partial differential equations problem - finding the general solution]]>

1. Which of the differential equations is/are non linear?

a. Only I

b. Only II

c. Only I and III

d. Only II and IV

e. None of the above

2. Which is a solution to the following DE? yy'-ye

a. y=t

b. y=...

Four assorted ODE questions]]>

r(t) is romeo's love for Juliet at time t, j(t) is Juliet's love for Romeo at time t

So far, it is given:

It is also given that Romeo & Juliet's families are enemies, thus the initial condition at time

If we would take the second derivative of r we get: r’’=-j’. We know that j’=r, which means r’’ =-r. can be recognized as the equation of...

Help: Differential equation Romeo & Juliet]]>

$\displaystyle

ty^\prime - 2y =\sin{t}, \quad t>0\\$

Divide thru by $t$

$\displaystyle y^\prime - \frac{2}{t}y =\frac{\sin{t}}{t}$

Obtain $u(t)$

$\displaystyle u(t)=\exp\int -

\frac{2}{t} \, dx =e^{2\ln{t}}=t^{-2}\\$

Multiply thru with $t^{-2}$

$t^{-2}y^\prime - 2 t y= t^{-2}\sin{t}\\$

Simplify:

$(t^{-2}y)'= t^{-2}\sin{t}\\$

Integrate:

$\displaystyle t^2y=\int t^{-2}\sin{t} dt =

2t\sin(t)-(t^2-2)\cos(t)+c_1\\$...

2.1.6 Another DE went off the rails]]>

\(\displaystyle x^2 + dy/dx + xy = 1\)

Can I just use algebra to do it? Like:

\(\displaystyle dy/dx = x^2/xy \)]]>

We have the initital value problem $$\begin{cases}y'(t)=1/f(t, y(t)) \\ y(t_0)=y_0\end{cases} \ \ \ \ \ (1)$$ where the function $f:\mathbb{R}^2\rightarrow (0,\infty)$ is continuous in $\mathbb{R}^2$ and continuously differentiable as for $y$ in a domain that contains the point $(t_0, y_0)$.

Show that there exists $h>0$ such that the following two conditions are satisfied:

- The problem (1) has a solution $\phi=\phi(t)$ that is defined at least for each $t\in (t_0-h...

Existence and uniqueness of solution]]>

I want to find the equilibrium solutions and determine their stability.

$(1)\left\{\begin{matrix}

\dot{x}=-y-x(1-\sqrt{x^2+y^2})^2\\

\dot{y}=x-y(1-\sqrt{x^2+y^2})^2

\end{matrix}\right.$

I also want to check the behavior of the solutions of $(1)$ when $t \to \infty$. There are four possible answers.

- there exists exactly one equilibrium solution and it is asymptotically stable. Furthermore, $\forall$ solution $(x,y)$ of $(1)$ we have that $\lim x(t)=x_0$ and...

Equilibrium solutions]]>

I want to solve the equation $u_t+uu_x=0$ with the initial condition $u(x,0)=1$ for $x \leq 0$, $1-x$ for $0 \leq x \leq 1$ and $0$ for $x \geq 1$. I want to solve it for all $t \geq 0$, allowing for a shock wave. I also want to find exactly where the shock is and show that it satisfies the entropy condition.

I have tried the following.

We get that $u(x(t),t)=c$.

The characteristic line that passes through the points $(x,t)$ and $(x_0,0)$ has slope...

Shock wave]]>

I want to show for the initial value problem of the wave equation

$$u_{tt}=u_{xx}+f(x,t), x \in \mathbb{R}, 0<t<\infty$$

that if the data (i.e. the initial data and the non-homogeneous term $f$) have compact support, then, at each time, the solution has also compact support.

I have thought the following:

The initial data are these, right?

$$u(x,0)=\phi(x) \\ u_t(x,0)=\psi(x)$$

The functions $f, \phi, \psi$ have compact support and so they are zero outside a bounded set...

Is this the desired bounded set of the wave equation?]]>

Let $\phi \in C^1(\mathbb{R})$ and periodic.

We consider the problem

$u_t=u_{xx}, x \in \mathbb{R}, \ 0<t<\infty$,

with initial data $\phi$.

I want to compute a formula for a solution $u$ and I want to prove strictly that this formula solves the initial value problem. I also want to show that there is no other solution ($C^1(\mathbb{R})$ and periodic).

I have thought the following.

$u(x,t)=X(x) T(t)$

$u(0,x)=\phi(x)$

$u_t=u_{xx} \Rightarrow X(x) T'(t)=X''(x) T(t)...

Find formula for solution heat equation]]>

I want to solve the Laplace equation on the unit disk, with boundary data $u(\theta)=\cos{\theta}$ on the unit circle $\{ r=1, 0 \leq \theta<2 \pi\}$. I also want to prove that little oscillations of the above boundary data give little oscillations of the corresponding solution of the Dirichlet problem after first stating strictly the statemtent.

How do we solve the Laplace equation on the unit disk with the given boundary data? Could you give me a hint? ]]>

$ \textsf{for which the given differential equation has solutions }$

$ \textsf{of the form $y = e^{rt}$}$

$\textit{15. $\quad y'+2y=0$}$

\begin{align*}

2\exp\int \, dx &=2e^{t}+c\\

&=e^{2t}+c\\

\therefore r&=2

\end{align*}

no sure about the rest due to mutiple differentiation

$\textit{16 $\quad y''-y=0$}$

$\textit{17. $\quad y''+y'-6y=0$}$

$\textit{18. $\quad y'''-3y''+2y'=0$}$]]>

We have the Cauchy problem of the equation

$u_t+xu_x=xu, x \in \mathbb{R}, 0<t<\infty$

with some given smooth ($C^1$) function $g$ as initial value.

I want to check if the problem is well defined for each time.

We know that a problem is well defined if the solution exists, is unique and depends continuously on the data of the problem.

I have computed that the solution of the problem is $u(x,t)=g(xe^{-t}) e^{x(1-e^{-t})}$.

So we have that the problem is...

Is it implied that the problem is well-defined?]]>

The eigenvalues of the matrix are $\lambda_1, \lambda_2 = 0,5$ and eigenvectors are $v_1 = (1,2)^t$ and $v_2 = (2, 1)^t$.

So I thought the solution would be $\mathbf{X} = c_1 (1,2)^t + c_2(2,1)^{t}e^{5t}$. But apparently this is wrong? Why?

P.S. the t on the vectors denotes...

Differential equation with a matrix]]>

\begin{align*}\displaystyle

\textit{if } r&=e^{t} \textit{then:}\\

y''+y'-2y&=r^2+r-2=(r+2)(r-1)=0\\

&=c_1^{2t}+c_2^{-t}\\

y&=\color{red} {e^t; \quad y \to \infty \,as \, t \to \infty }

\end{align*}

ok the bk answer is in red

I just don't get this

why do we need $$y(0)=1 \quad y'(0)=1$$

the graph of $y=e^t$

]]>

find the general solution of the second order differential equation.

$$y''+2y'-3y=0$$

assume that $y = e^{rt}$ then,

$$r^2+2r-3=0\implies (r+3)(r-1)=0$$

new stuff.... so far..]]>

$\textsf{ Solve each of the following initial value problems and plot the solutions for several values of $y_0$.}\\$

$\textsf{ Then describe in a few words how the solutions resemble, and differ from, each other.}\\$

$$\begin{align*}\displaystyle

\frac{dy}{dt}&=-y+5, \quad y(0)=y_0\\

y'&=-y+5\\

y'+y&=5\\

u(t)&=\exp\int \, dt = e^t\\

e^ty'+e^ty&=e^t5\\

(e^ty)'&=e^t5\\

e^ty &=5\int e^t dt=5e^t+c\\

y...

de1.2.1 Solve the following initial value problem]]>

$$y''-2y=e^{2t} \quad y(0)=2$$

not sure about the $e^{2t}$]]>

$\textsf{Given: }$

$$\displaystyle y'=2xy^2, \quad y(0)=1$$

$\textit{Show that $y=\phi(x)=(1-x^2)^{-1}$ is a solution of the initial value problem}$

\begin{align*}\displaystyle

y'&=2xy^2\\

\frac{dy}{y^2}&=2x \\

y&=\color{red}{\frac{1}{(c_1-x^2)}}

\end{align*}

ok I went into confusion after 2x??

red is W|F]]>

Show that the general solution of the differential equation

y″(x)=−y(x)

is

y(x)=Acos(x)+Bsin(x)

where A and B are arbitrary constants. Hint: You'll need the Taylor series representations for sin and cos.

]]>

$\textsf{Find the solution of the given initial value problem}$

$$y'+2y=te^{-2t} \quad y(1)=0$$

$\textit{obtain $u(x)$}$

$$\exp\int 2 dt=e^{2t}$$

$\textit{multiply thru by $e^{2t}$}$

$$e^{2t}y'+2e^{2t}y=(e^{2t}y)'=t$$

$\textit{Integrate}$

$$e^{2t}y= \frac{t^2}{2} + c_1$$

$\textit{Divide thru by $e^{2t}$ }$

$$y= \frac{t^2}{2e^{2t}} + \frac{c_1}{e^{2t}}$$

$\textit{W|A}$

$ \color{red}{y(t) =c_1e^{-2t}+\frac{1}{2}e^{-2t}t^2 }$

not sure about intial value]]>

$$y'=\frac{x^2}{y}$$

ok this is a new section on separable equations

so i barely know anything

but wanted to post the first problem

hoping to understand what the book said.

thanks ahead...]]>

$$y'= \frac{x^2}{y(1+x^3)}$$

Multiply both sides by the denominator

$$y(1+x^3)y'=x^2$$

Subtract $x^2$ from both sides

$$-x^2 +y(1+x^3)y'=0$$

ok was trying to follow an example but ???

View attachment 8242]]>

$\textsf{Find the solution of the given initial value problem}$

$$y'-y=2te^{2t}, \quad y(0)=1$$

$\textit{Find u(x)}$

$$\displaystyle\exp\int -1 dt =e^{ t^{-1}}$$

$\textit{multiply thru with $e^{ t^{-1}}$} $

$$ e^{ t^{-1}}y'- e^{ t^{-1}}y=2te^{t}$$

ok this isn't uv'+u'v

$\textit{W|A}$

$$\color{red}{c_1e^t+2e^{2t}t-2e^{2t}}$$]]>

$\displaystyle ty^\prime -y =t^2e^{-1}\\$

Divide thru by t

$\displaystyle y^\prime-\frac{y}{t}=te^{-1}\\$

Obtain $u(t)$

$\displaystyle u(t)=\exp\int -\frac{1}{t} \, dt =\frac{1}{t}\\$

Multiply thru with $\displaystyle\frac{1}{t}$

$\displaystyle y^\prime -\frac{1}{t}y =te^{-1}\\$

Simplify:

$(\frac{1}{t}y)'=te^{-1} \\$

ok got stuck here so...

Answer from bk

$\displaystyle

\color{red}{y=c_1 -te^{-1}}$...

2.1.10 ty' -y =t^2e^{-1} u(x)???]]>

$\displaystyle y^\prime -2ty =2te^{-t^2}\\$

Obtain $u(t)$

$\displaystyle u(t)=\exp\int -2t \, dt =e^{-t^2}$$%e^(-t^2)\\$

Multiply thru with $e^{-t^2}$

$(e^{-t^2})y^\prime -(e^{-t^2})2ty =(e^{-t^2})2te^{-t^2}\\$

Simplify:

$((e^{-t^2})y)'= 2te^{-2t^2}\\$

Integrate:

$\displaystyle e^{-t^2}y=\int 2te^{-2t^2} dt =-\frac{ e^{-2t^2}}{2}+c_1\\$

Divide by $e^{-t^2}$

$\displaystyle y=-\frac{e^{-t^2}}{2}+c_1 e^{t^2}\\$...

2.1.7 DE y'e -2ty =2te^(-t^2)]]>