Ivp Definition and 144 Threads

My code is as follows: but when I use the function in my command window exactsol(t) and input a tolerance but there is an error in LINE 19 saying unrecognized ivpfun, could someone help me fix it as I am unsure of how to proceed from here. function y = exactsol(t) y = zeros (2,1); y(1) =...

\[ y'=y+2te^{2t} \]

Find the solution of $y^{\prime} = \dfrac{x(x^2+1)}{4y^3}, \quad y(0)=-\dfrac{1}{\sqrt{2}}$ in explicit form. rewrite $\dfrac{y}{x}=\dfrac{x(x^2+1)}{4y^3}= y^3\, dy=\dfrac{x(x^2+1)}{4}\, dx$ integrate $\dfrac{y^4}{4}= \dfrac{1}{4}\left(\dfrac{x^4}{4} +\dfrac{ x^2}{2}\right)$...

find the solution of the IVP $ty'+2y=t^2-t+1, \quad y(1)=\dfrac{1}{2}, \quad t>0$\begin{array}{lll} \textsf{Divide thru by t} & y'+\dfrac{2}{t}y=t-1+\dfrac{1}{t}\\ \textsf{Find u(t)} &u(t)=\exp\displaystyle\int \dfrac{2}{t} \, dx =e^{2ln |t|}+c \end{array} so far anyway... is...

$\tiny{b.1.3.7}$ Solve IVP $y''-y=0;\quad y_1(t)=e^t,\quad y_2(t)=\cosh{t}$ $\begin{array}{lll} &\exp\left(\int \, dx\right)= e^x\\ & e^x(y''-y)=0\\ & e^x-e^x=0\\ \\ &y_1(x)=e^x\\ &(e^x)''-(e^x)=0\\ &(e^x)-(e^x)=0\\ \\ &y_2(x)=\cosh{x}\\ &(\cosh{x})''-(\cosh{x})=0\\ \end{array}$ ok there was...

Solve IVP $\begin{array}{rl}x' & = 2x + 2y\\y' & = -4x + 6y\\x(0) & = 2\\y(0) & = -3 \end{array}$ assume we can proceed with this first $A=\left[\begin{array}{rr}2&2\\-4&6\end{array}\right]\\ A-rI=\left[\begin{array}{rr}2-r&2\\-4&6-r\end{array}\right]=r^{2} -8r + 20 = 0 \quad r_1 = 4 -2 i \quad...

ck for typos https://photos.app.goo.gl/eRfYNAVK1jnBgSCu8 https://photos.app.goo.gl/8C9sJ9UgZbxXgP4P9 Boyce Book (a) Transform the given system into a single equation of second order. (b) Find $x_1$ and $x_2$ that also satisfy the given initial conditions. (c) Sketch the graph of the solution...

$\tiny{2.1.5.1.c}$ source Change the second-order IVP into a system of equations $\dfrac{d^2x}{dt^2}+\dfrac{dx}{dt}'+4x=\sin t \quad x(0)=4\quad x'(0)= -3$ ok I presume we can rewrite this as $u''+u'+4u=\sin t$ Let $x_1=u$ and $x_2=u'$ then $x_1'=x_2$ substituting $x_2'+x_2+4x=\sin t$...

Change the second-order IVP into a system of equations $y''+y'-2y=0\quad y(0)= 2\quad y'(0)=0$ let $x_1=y$ and $x_2=y'$ then $x_1'= x_2$ and $y''=x_2'$ then by substitution $x_2'+x_2-2x_1=0$ then the system of first order of equations $x_1'=x_2$ $x_2'=-x_2+2x_1$ hopefully so far..

source Change the second-order IVP into a system of equations $y''+y'-2y=0 \quad y(0)= 2\quad y'(0)=0$ let $u=y'$ ok I stuck on this substitution stuff

$\tiny{2.1.5.1}$ Change the second-order initial-value problem into a system of equations $x''+6x'-2x= 0\quad x(0)=1\quad x'(0)=1$ ok my first step was to do this $e^{rt}(r^2+6r-2)=0$ using quadratic formula we get $r=-3+\sqrt{11},\quad r=-3-\sqrt{11}$ just seeing if I going down the right road🕶

Summary:: Initial value problem Solve the given initial-value problem differential equation by undetermined coefficient method. d^2 x/dt^2 + w^2 x = F sin wt , x(0) = 0, x'(0) = 0 I get the sol = C1 cos wt + C2 sin wt, but i always get 0 when I plug into the equation, anyone can help me...

$y'=ty\dfrac{4-y}{3};\quad y{0}=y_0$ separate $\dfrac{3}{y(4-y)}\ dy=t\ dt$ integrate $3\left(\frac{1}{4}\ln \left|y\right|-\frac{1}{4}\ln \left|y-4\right|\right)=\dfrac{t^2}{2}+c$ hopefully so far... actually it kinda foggy what they are eventually asking for also why is t in different cases

\tiny{b.48.2.2.23} Solve the initial value problem $y'=2y^2+xy^2,\quad y(0)=1$ find min value $\begin{array}{ll} \textit{separate variables} &\dfrac{1}{y^{2}}\ dy=(2+x)\ dx\\ \\ \textit{integrate} & -\dfrac{1}{y}=2x+\dfrac{x^2}{2}+C\\ \\ \textit{plug in...

368 Solve the IVP $y'=2(1+x)(1+y^2),\quad y(0)=0$ $\begin{array}{ll} \textit{separate variables}& \displaystyle \left(\dfrac{1}{1+y^2}\right)\ dy =2(1+x)\ dx\\ \textit{integrate thru}& \arctan \left(y\right)=2x+x^2+c\\ \textit{plug in x=0 and y=0}& \arctan 0=0+c\\ &0=c\\ \textit{thus the...

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 \[(3+2y) \, dy= (2\cos 2x) \, dx\] $y^2 + 3 y= sin(2 x) + c$

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...

2000 given #23 so far I could not get to the W|A solution before applying the y(0)=ahere is the book answer for the rest

(a) find solution of initial value and (c) interval $$\quad\displaystyle y^2(1-x^2)^{1/2} \,dy = \arcsin{x}\,dx, \quad y(0) = 1$$ 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...

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

693 2.2.13 (a) find initial value (b)plot and (c) interval (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...

$$\d{r}{\theta}=\frac{r^2}{\theta},\quad r(1)=2$$ 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

\nmh{2000} find the solution of the given initial value problem $$y''-2y=e^{2t} \quad y(0)=2$$ not sure about the $e^{2t}$

$\tiny{2.3.2}$ 1000 $\textsf{Given: }$ $$\displaystyle y'=2xy^2, \quad y(0)=1$$ $\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)}}...

$\tiny{b.2.1.16}$ \begin{align*}\displaystyle y^\prime +\frac{2}{x}y &=\frac{\cos x}{x^2} &&(1)\\ u(x)&=\exp\int\frac{2}{x} \, dx = e^{2\ln{x}}=2x&&(2)\\ (2x y)'&=\int\frac{\cos x}{x^2} \, dx &&(3)\\...

Hello! (Wave) I want to prove that if the initial data of the initial value problem for the wave equation have compact support, then at each time the solution of the equation has also compact support. Doesn't the fact that a function has compact support mean that the function is zero outside...

Hi PF! I'm reading a text where the authors construct a Green's function for a given BVP by variation of parameters. The authors construct the Green's function by finding first the fundamental solutions (let's call these ##v_1## and ##v_2##) to the homogenous BVP. However, the authors determine...

Hi PF! Generally speaking, how would one solve $$f''-a(x) f = 0 : f(0)=0,f'(0)=1$$ Or if you could point me to a source that would be awesome too!

$\tiny{2214.t1.13}$ 1000 $\textsf{solve the initial value problem}$ \begin{align*}\displaystyle \frac{ds}{dt}&=12t(3t^2-1)^3 ,s(1)=3\\ I_{13}&=\int 12t(3t^2-t)^3 \, dt\\ u&=(3t^2-1) \therefore \frac{1}{6t}du=dt \\ &=2\int u^3 du\\...

Homework Statement [/B] It's been a couple of years since differential equations so I am hoping to find some guidance here. This is for numerical analysis. Any help would be much appreciated. Homework EquationsThe Attempt at a Solution

Homework Statement I need to solve: x^2y''-4xy'+6y=x^3, x>0, y(1)=3, y'(1)=9 Homework EquationsThe Attempt at a Solution I know that the answer is: y=x^2+2x^3+x^3lnx Where did I go wrong. I was wondering if it's even logical to solve it as an Euler Cauchy and then use variation of parameters...

I am being asked to find ##\lambda## such that ##y'' + \lambda y = 0; ~y(0) =0; ~y'( \pi ) = 0##. This is an eigenvalue problem where we are given boundary conditions. ##\lambda## can be found such that we don't have a trivial solution if we test the different cases when ##\lambda < 0, \lambda =...

Homework Statement Use laplace transforms to find following initial value problem -- there is no credit for partial fractions. (i assume my teacher is against using it..) y'' + 2y' + 2y = 2 ; y(0)= y'(0) = 0 Homework Equations Lf'' = ((s^2)*F) - s*f(0) - f'(0) Lf' = sF - f(0) Lf = F(s) The...

x' =-5x-y y' =4x-y I got x=ae^-3t+bte^-3t y=-2bte^-3t+2ae^-3t+be^-3t a=0 b=0 The answer is x=e^(-3t+3)-te^(-3+3) y=-e^(-3+t)+2te^(-3+3) I don't understand where the +3 comes from

Homework Statement (didn't know how to make piecewise function so I took screenshot) Homework EquationsThe Attempt at a Solution My issue here with this problem is that I have absolutely no idea where to start... I have read through the textbook numerous times, and searched all over the...

Hey! :o If we have the initial and boundary value problem $$u_{tt}(x,t)-c^2u_{xx}(x,t)=0, x>0, t>0 \\ u(0,t)=0 \\ u(x,0)=f(x), x\geq 0 \\ u_t(x,0)=g(x)$$ and we want to apply Green's theorem do we have to expand the problem to $x \in \mathbb{R}$ ?? (Wondering)

Ok so for this problem I have to use IVPs to prove that cos^2(x)+sin^2(x)=1. I know the end result is suppose to be: du/dt= - v, u(0)=1 dr/dt= u, v(0)=0 but I have no idea how to go about getting to this point.

When the Laplace Transform is applied to a certain IVP, the resulting equation is $(s^2Y - 2s - 1) + (sY - 2) - 2Y = \frac{4}{s}$ What was the IVP? So I think I've solved this, but just want to make sure I got the correct answer. I got: $y(0) = 2$ , $y'(0) = 1$, & $C = 2$ $\therefore y'' +...

Need someone to verify that my solution is correct, thanks in advance. Solve the IVP $y'' - y = e^t$, $y(0) = 0$, $y'(0) = 1$ Solution: $\frac{1}{2}te^t + \frac{1}{2}e^t - \frac{1}{2}e^{-t}$

\frac {dy}{dt} = y^3 + t^2, y(0) = 0 My teacher said this IVP couldn't be expressed in terms of functions we commonly know. I was wondering what the general solution is? Thank-you

Find the unique solution to the IVP $t^3y'' + e^ty' + t^4y = 0$ $y(1) = 0$ , $y'(1) = 0$ Should I start out by dividing through by $t^4$ to get $\frac{1}{t} y" + \frac{e^t}{t^4}y' + y = 0$

Solve the IVP. $(2x-y)dx + (2y-x)dy = 0 $. $y(1) = 3$. Leave solution in implicit form. So I got: $\frac{dy}{dx} = \frac{-(2x-y)}{2y-x}$ Would this be correct since I didn't explicitly solve for $dy$ ?

Find the explicit solution to the IVP. $xdx + ye^{-x}dy=0$, $y(0) =1$ so I did some manipulation to get $ye^{-x}dy= -xdx$ ==> $\frac{dy}{dx}=\frac{-x}{ye^{-x}}$ but now I'm confused on what to do. What I found above is the implicit solution right? So do I just need to get $y'$ on the left side...

Homework Statement http://s14.postimg.org/an6f4t2ht/Untitled.png Homework EquationsThe Attempt at a Solution I'm not sure what they want me to do on the last part. I tried some googling and looking in my textbooks but I didn't find any examples. It seems to me like the function goes to...

Homework Statement Homework Equations DifEqs The Attempt at a Solution y ' = 4C1e-4xSinX - 4C2e-4xCosX y'(0) = -1 -1 = 0 - 4C2 Therefore C2 = 1/4 Not correct. What am I doing wrong?

Homework Statement Solve the IVP : dy/dt + y = f(t) y(0) = -5 where f(t) = -1, 0 <= t < 7 -5, t >= 7 y(t) for 0 <= t < 7 = ? y(t) for t >= 7 = ? Homework EquationsThe Attempt at a Solution So I have never seen a problem of this type, excuse my silly mistakes if I'm interpreting this question...

x=c_1 cos(t)+c_2sin(t) is a family of solution to x''+x=0. Given x(\frac{\pi}{6})=\frac{1}{2} and x'(\frac{\pi}{6})=0 find a solution to the second order IVP consisting of this differential equation and the given intial conditions. The answer key has x=\frac{\sqrt{3}}{4}cos(t)+\frac{1}{4}sin(t)...

Are these statements correct, if not could you give me an example 1. If solution of IVP is non-unique then there are infinitely many solutions in short, if the solution to the IVP has at least 2 solutions then there are infinitely many solutions to this IVP 2.there are none IVP first...