Question

who can help me answering differential equation?

Answer 1

Do not ask to ask.

Answer 2

given initial value problem x(alpha)=beta

Answer 3

\[(\cos t) x'=(\sin t) x-\sqrt{t-1}\]

Answer 4

Subtact \((\sin t)x\) from both sides so we have \((\cos t)x'-(\sin t)x=-\sqrt{t-1}\). Note our left-hand side is just the derivative of \(x\cos t\). Integrate both sides:$$x\cos t=-\int\sqrt{t-1}\,\mathrm{d}t=-\int\sqrt{u}\,\mathrm{d}u\text{ where }u=t-1$$I trust you can take it from here.

Answer 5

$$x\cos t=-\frac23u^\frac32+C=-\frac23(t-1)^\frac32+C\\x(t)=-\frac{2\sqrt{(t-1)^3}}{3\cos t}+C=-\frac{2\sqrt{t^3-3x^2+3x-1}}{3\cos t}+C$$Use your initial condition to solve for \(C\).

Answer 6

when i integrate i get 1/tanx = 3/2 (t-1)^3/2 +c..what wrong with my answer?

Answer 7

Did you try solving via an integration factor \(\mu=e^{-\int\tan x\,\mathrm{d}x}\)? There's no need -- we can skip steps.

Answer 8

\[\int\limits_{}^{} \frac{ d }{ dt}(\frac{ \cos t }{ \sin t}) dt = -\int\limits_{}^{} \sqrt{t-1}dt\]

Answer 9

oops I broke my solution. It should be $$x(t)=-\frac{2\sec t}3\left((t-1)^\frac32+C\right)$$

Answer 10

@aiskarl I'll solve it that way in a second.

Answer 11

$$(\cos t)x'=(\sin t)x-\sqrt{1-t}\\(\cos t)x'-(\sin t)x=-\frac{\sqrt{1-t}}{\cos t}\\x'-(\tan t)x=-\sqrt{1-t}$$so we pick an integration factor \(\mu=e^{\int-\tan t\mathrm\,{d}t}=e^{\log(\cos t)}=\cos t\) and multiply both sides:$$(\cos t)x'-(\sin t)x=-\sqrt{1-t}$$... hey, isn't that exactly what we had? this is what I meant by skipping steps -- we didn't need an integration factor and could've used the product rule immediately.
We know \((\cos t)x'-(\sin t)x=(x\cos t)'\) so we rewrite:$$(x\cos t)'=-\sqrt{1-t}\\\int(x\cos t)'=\int-\sqrt{1-t}\\x\cos t=-\int\sqrt{1-t}=-\frac23(t-1)^\frac32+C\\x(t)=-\frac2{3\cos t}\left((t-1)^\frac32+C\right)=-\frac{2\sec t}3\left(\sqrt{t^3-3t^2+3t-1}+C\right)$$

Answer 12

wait ..i cannot understand..start from
... hey, isn't that exactly what we had? this is what I meant by skipping steps -- we didn't need an integration factor and could've used the product rule immediately.

Answer 13

how we know\[(\cos t)x'-(sint )x=(xcos t)'\]

Answer 14

implicit differentiation :)

Answer 15

not reaally i think..did you know about product rule..i'm trying to do product rule

Answer 16

product rule, but differentiate x implicitly

Answer 17

i cannot understand..could u help me

Answer 18

@aiskarl \(\frac{d}{dt}(x\cos t)=x\frac{d}{dt}\cos t+\cos t\frac{d}{dt}x=-x\sin t+x'\cos t=(\cos t)x'-(\sin t)x\)

Answer 19

Teorem 1.1 (Linear Case)
If the functions \[a_{0}(x),a _{1}(x),...,a _{n}(x) and g(x)\]are continuous on the interval I and \[a _{0 }(x) \neq 0,\forall x \notin I\], then the initial value problem
\[a _{0}(x) y ^{(n)}+a _{1}(x)y ^{(n - 1)}+...+a _{n}(x)y=g(x)\]
\[y(x _{0})=y _{0},y'(x _{0})=y _{1},...,y ^{(n-1)}(x _{0})=y _{n-1},x _{0}\in I\]
has a unique solution on I.

Answer 20

Theorem 1.2 : Existence of a Unique Solution (General case)
Consider the initial value problem \[\frac{ dy }{ dx }=f(x,y) , y(x _{0})=y _{0}\]
Let R be a rectangular region in xy plane defined by ,a≤x≤b,c≤y≤d that contains the point \[(x _{0},y _{0})\]in its interior. If f (x,y) and \[ \frac{ \delta f }{ \delta y }\]are continuous on R, then there exists some interval \[I _{0}:x _{0}-h<x,x _{0}=h,h>0\]contained in a≤x≤b, and a unique function y(x), defined on \[I _{0}\], that is a solution of the given initial value problem.

Answer 21

continued from question..
(i) given \[\alpha = 1.5, \beta=0\] find the interval so that the initial value theorem (ivp) is guaranteed by existance and uniqueness theorem (EUT) to have a unique solution by using the following three method
a) theorem 1.1 with \[a _{0}(t)=\cos t,where a _{0}\] is coefficient of x'
b)theorem 1.1 with \[a _{0}(t) =1\] after modifying DE
c)theorem 1.2

Answer 22

?

Answer 23

find the interval

Answer 24

so we have \(x(\frac32)=0\), which means \((t-1)^\frac32+C=0\). Here's pretty clear that \(C=-(\frac12)^\frac32=\frac1{2\sqrt2}=\frac{\sqrt2}{4}\)

Answer 25

interval of existness/uniqueness??

Answer 26

Well we have a linear differential equation here so Theorem 1.1 is the existence-uniqueness theorem we want. We know our interval must contain our point given as an initial condition \((\frac32,0)\) and we need \(\cos t\), \(\sin t\), and \(\sqrt{1-t}\) to be continuous... hmmm :-)

Answer 27

Well, \(\sin x\) and \(\cos x\) are continuous everywhere but \(\sqrt{1-t}\) is continuous on the interval \((-\infty,1)\). I'm going to hazard a guess that the interval for (a) is \((-\infty,1)\).

For (b), I imagine it's the same.