manierato5h

2022-06-29

The theorem of existence and uniqueness is: Let ${y}^{\prime}+p(x)y=g(x)$ be a first order linear differential equation such that p(x) and g(x) are both continuous for $a<x<b$. Then there is a unique solution that satisfies it.

When a differential equation has no solution that satisfies $y({x}_{0})={y}_{0}$, what does this mean?? Can the theorem be verified??

When a differential equation has no solution that satisfies $y({x}_{0})={y}_{0}$, what does this mean?? Can the theorem be verified??

scipionhi

Beginner2022-06-30Added 25 answers

The existence and uniqueness theorem for first-order linear differential equations can be stated as follows. Suppose that P and Q are continuous on the open interval I. If a and b are any real numbers, then there is a unique function y=f(x) satisfying the initial-value problem ${y}^{\prime}+P(x)y=Q(x)$ with f(a)=b on the interval I. With regard to your question, the important point is that a and b are arbitrary real numbers and the unique solution f to the differential equation satisfies f(a)=b for every choice of a and b. Since every first-order linear differential equation satisfying the constraints of the theorem has a solution satisfying f(a)=b, there is no case in which such an equation has no solution satisfying f(a)=b.

If we look at the simpler case of homogeneous first-order linear differential equations of the form ${y}^{\prime}+P(x)y=0$, where P is continuous on the open interval I, we can directly verify that for every choice of a and b, the function $f(x)=b{e}^{-A(x)}$ where $A(x)={\int}_{a}^{x}P(t)dt$ is a solution to ${y}^{\prime}+P(x)y=0$. Now letting g be an arbitrary solution of ${y}^{\prime}+P(x)y=0$, we establish uniqueness by showing that $g(x){e}^{A(x)}=b$. Differentiating, we see that $h(x)=g(x){e}^{A(x)}$ is constant on the interval I. But h(a)=b, so we must have h(x)=b. This demonstrates that g=f. Notice that the choice of a and b does not affect the existence or uniqueness of solutions. Can you verify this in the case of non-homogeneous first-order linear differential equations of the form ${y}^{\prime}+P(x)y=Q(x)$

If we look at the simpler case of homogeneous first-order linear differential equations of the form ${y}^{\prime}+P(x)y=0$, where P is continuous on the open interval I, we can directly verify that for every choice of a and b, the function $f(x)=b{e}^{-A(x)}$ where $A(x)={\int}_{a}^{x}P(t)dt$ is a solution to ${y}^{\prime}+P(x)y=0$. Now letting g be an arbitrary solution of ${y}^{\prime}+P(x)y=0$, we establish uniqueness by showing that $g(x){e}^{A(x)}=b$. Differentiating, we see that $h(x)=g(x){e}^{A(x)}$ is constant on the interval I. But h(a)=b, so we must have h(x)=b. This demonstrates that g=f. Notice that the choice of a and b does not affect the existence or uniqueness of solutions. Can you verify this in the case of non-homogeneous first-order linear differential equations of the form ${y}^{\prime}+P(x)y=Q(x)$

- A body falls from rest against resistance proportional to the square root of the speed at any instant. If the body attains speed V1 and V2 feet per second, after 1 and 2 seconds in motion, respectively, find an expression for the limiting velocity.

The Laplace transform of the product of two functions is the product of the Laplace transforms of each given function. True or False

The Laplace transform of

is$u(t-2)$

(a)$\frac{1}{s}+2$

(b)$\frac{1}{s}-2$

(c) ??$e}^{2}\frac{s}{s}\left(d\right)\frac{{e}^{-2s}}{s$ 1 degree on celsius scale is equal to

A) $\frac{9}{5}$ degree on fahrenheit scale

B) $\frac{5}{9}$ degree on fahrenheit scale

C) 1 degree on fahrenheit scale

D) 5 degree on fahrenheit scaleThe Laplace transform of $t{e}^{t}$ is A. $\frac{s}{(s+1{)}^{2}}$ B. $\frac{1}{(s-1{)}^{2}}$ C. $\frac{s}{(s+1{)}^{2}}$ D. $\frac{s}{(s-1)}$

What is the Laplace transform of

into the s domain?$t\mathrm{cos}t$ Find the general solution of the given differential equation:

${y}^{\u2033}-2{y}^{\prime}+y=0$The rate at which a body cools is proportional to the difference in

temperature between the body and its surroundings. If a body in air

at 0℃ will cool from 200℃ 𝑡𝑜 100℃ in 40 minutes, how many more

minutes will it take the body to cool from 100℃ 𝑡𝑜 50℃ ?A body falls from rest against a resistance proportional to the velocity at any instant. If the limiting velocity is 60fps and the body attains half that velocity in 1 second, find the initial velocity.

What's the correct way to go about computing the Inverse Laplace transform of this?

$\frac{-2s+1}{({s}^{2}+2s+5)}$

I Completed the square on the bottom but what do you do now?

$\frac{-2s+1}{(s+1{)}^{2}+4}$How to find inverse Laplace transform of the following function?

$X(s)=\frac{s}{{s}^{4}+1}$

I tried to use the definition: $f(t)={\mathcal{L}}^{-1}\{F(s)\}=\frac{1}{2\pi i}\underset{T\to \mathrm{\infty}}{lim}{\int}_{\gamma -iT}^{\gamma +iT}{e}^{st}F(s)\phantom{\rule{thinmathspace}{0ex}}ds$or the partial fraction expansion but I have not achieved results.How do i find the lapalace transorm of this intergral using the convolution theorem? ${\int}_{0}^{t}{e}^{-x}\mathrm{cos}x\phantom{\rule{thinmathspace}{0ex}}dx$

How can I solve this differential equation? : $xydx-({x}^{2}+1)dy=0$

Find the inverse Laplace transform of $\frac{{s}^{2}-4s-4}{{s}^{4}+8{s}^{2}+16}$

inverse laplace transform - with symbolic variables:

$F(s)=\frac{2{s}^{2}+(a-6b)s+{a}^{2}-4ab}{({s}^{2}-{a}^{2})(s-2b)}$

My steps:

$F(s)=\frac{2{s}^{2}+(a-6b)s+{a}^{2}-4ab}{(s+a)(s-a)(s-2b)}$

$=\frac{A}{s+a}+\frac{B}{s-a}+\frac{C}{s-2b}+K$

$K=0$

$A=F(s)\ast (s+a)$