What is Ivp: Definition and 144 Discussions

InterVarsity Press (IVP) was founded in 1947 by InterVarsity Christian Fellowship/USA as a publisher of evangelical Christian books. It is headquartered in Downers Grove, Illinois.

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  1. V

    Solving An IVP on Matlab with ODE 45 with different tolerances

    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) =...
  2. H

    MHB Find the solution y to the IVP

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

    MHB B.2.2.16 IVP of \dfrac{x(x^2+1)}{4y^3}, y(0)=-\dfrac{1}{\sqrt{2}}

    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 $\dfrac{y^4}{4}= \dfrac{1}{4}\left(\dfrac{x^4}{4} +\dfrac{ x^2}{2}\right)$...
  4. karush

    MHB What is the integrating factor for solving this IVP?

    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...
  5. karush

    MHB Solving IVP $y''-y=0$ with $y_1,y_2$

    $\tiny{b.1.3.7}$ 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...
  6. karush

    MHB Solve IVP: | Eigenvectors Found

    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...
  7. karush

    MHB -7.1.8 IVP with system of eq

    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...
  8. karush

    MHB Solving 2nd-Order IVP as System of Equations

    $\tiny{}$ 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 $x_2'+x_2+4x=\sin t$...
  9. karush

    MHB System of Equations for Second-Order IVP

    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..
  10. karush

    MHB Converting a Second-Order IVP into a System of Equations: Can Substitution Help?

    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
  11. karush

    MHB 097 Change the second-order IVP into a system of equations

    $\tiny{}$ 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🕶
  12. Fallen6

    Solve IVP: Undetermined Coefficient Method

    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...
  13. karush

    MHB What is the solution for t when $-0.1143e^{-\frac{2t^2}{3}}= -199$?

    $y'=ty\dfrac{4-y}{3};\quad y{0}=y_0$ 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
  14. karush

    MHB IVP and min valuew

    \tiny{b.} 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...
  15. karush

    MHB -b.2.2.26 IVP min value y'=2(1+x)(1+y^2); y(0)=0

    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...
  16. karush

    MHB -2.2.27 - Analysis of first order IVP

    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
  17. karush

    MHB -b.2.2.26 Solve first order IVP and determine where minimum of solution occurs

    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 $$(1+y^2)\, dy=(2+2x)\, dx$$
  18. karush

    MHB What is the Solution to the Differential Equation dy/dx = (2cos 2x)/(3+2y)?

    it's late so I'll just start this \[ \dfrac{dy}{dx}=\dfrac{2\cos 2x}{3+2y} \] so \[(3+2y) \, dy= (2\cos 2x) \, dx\] $y^2 + 3 y= sin(2 x) + c$
  19. karush

    MHB Can You Solve This Initial Value Problem and Determine Its Solution Interval?

    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...
  20. karush

    MHB Solve IVP 2000 #23: Y(0)=A Solution

    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
  21. karush

    MHB -2.2.20 IVP interval....trig subst y^2(1-x^2)^{1/2} \,dy=\arcsin{x}\,dx

    (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...
  22. karush

    MHB -b.2.2.18 IVP DE complete the square?

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

    MHB -2.2.12 IVP y'=2x/(y+x^2y), y(0)=-2

    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...
  24. karush

    MHB -2.2.12 dr/dθ = r^2/θ, r(1) = 2 IVP, graph, interval

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

    MHB -2.1.17 IVP y''-2y=e^{2t} \quad y(0)=2

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

    MHB -aux.2.2.02 y=\phi(x)=(1-x^2)^{-1} is a solution IVP

    $\tiny{2.3.2}$ 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)}}...
  27. karush

    MHB Solve IVP: What is the solution to the given initial value problem?

    $\tiny{b.2.1.16}$ \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)\\...
  28. evinda

    MHB Compact Support of Wave Equation IVP Solutions

    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...
  29. M

    A How do we go from BVP to IVP in determining the Green's function?

    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...
  30. M

    A Solving linear 2nd order IVP non-constant coefficient

    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!
  31. karush

    MHB -t1.13 IVP \frac{ds}{dt}&=12t(3t^2-1)^3 ,s(1)=3\

    $\tiny{2214.t1.13}$ 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\\...
  32. R

    Efficient Solutions for IVP and Root Approximation in Differential Equations

    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
  33. Cocoleia

    Second order non homogeneous ODE, IVP

    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...
  34. Mr Davis 97

    I BVP vs IVP when solving for eigenvalues

    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 =...
  35. Aristotle

    Laplace Transform Method for Solving Initial Value Problems

    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...
  36. M

    Why Does the Solution to My Differential Equation Include a +3?

    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
  37. Destroxia

    Step Function IVP Differential Equation w/ Laplace Transform

    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...
  38. M

    MHB Solving an IVP with Green's Theorem: Wondering?

    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)
  39. S

    Prove cos^2(x)+sin^2(x)=1 using IVP's

    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.
  40. S

    MHB Yes, that is the correct answer. Good job!

    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'' +...
  41. S

    MHB Verify My IVP Help Appreciated!

    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: $\frac{1}{2}te^t + \frac{1}{2}e^t - \frac{1}{2}e^{-t}$
  42. L

    What's the General Solution to This IVP?

    \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
  43. S

    MHB Find the unique solution to the IVP

    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 $\frac{1}{t} y" + \frac{e^t}{t^4}y' + y = 0$
  44. S

    MHB Solving the IVP, leaving it in Implicit Form

    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 $dy$ ?
  45. S

    MHB Finding the explicit solution to the IVP

    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...
  46. Feodalherren

    Solving IVPs with Unstable Functions

    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...
  47. Feodalherren

    Find the constants for given IVP

    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?
  48. T

    Laplace Transform for Solving a First Order Linear IVP

    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...
  49. F

    MHB Find specific solution to IVP with two parameters

    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)...
  50. K

    Problems with uniqueness of IVP

    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...