Can I say that

Magdalena Norton
2022-05-02
Answered

Can I say that

You can still ask an expert for help

timbreoizy

Answered 2022-05-03
Author has **15** answers

Step 1

$y=-1$ is not a singular solution because it is included in the general solution $y=C{e}^{x}-1$ when $C=0$ .

Solutions are only called singular when they are not attainable from the general solution form. In fact, I believe all linear first order homogenous equations have no singular solutions.

Solutions are only called singular when they are not attainable from the general solution form. In fact, I believe all linear first order homogenous equations have no singular solutions.

Giancarlo Brooks

Answered 2022-05-04
Author has **11** answers

Step 1

$y=-1$ is not a singular solution, because it does not pass by any point $({x}_{0},\text{}{y}_{0})$ such that the initial value problem

$\{\begin{array}{l}{y}^{\prime}=y+1\\ y({x}_{0})={y}_{0}\end{array}$

has more than one solution: in point of fact, the Cauchy problem

$\{\begin{array}{l}{y}^{\prime}=y+1\\ y({x}_{0})=-1\end{array}$

has only the one maximal solution$y=-1$ , whatever $x}_{0$ is.

has more than one solution: in point of fact, the Cauchy problem

has only the one maximal solution

asked 2021-01-02

Find

asked 2022-04-21

Variation of parameters method for differential equations.

Change the variable$x={e}^{t}$ and then find the general solution for the following differential equation $2{x}^{2}y{}^{\u2033}-6x{y}^{\prime}+8y=2x+2{x}^{2}\mathrm{ln}x$

It seems a little suspicious that i can factor out 2 and x first.${x}^{2}y{}^{\u2033}-3x{y}^{\prime}+4y=x(\mathrm{ln}\left(x\right)+1)$ and if we factor out $x}^{2$ now we end up with $y{}^{\u2033}-\frac{3}{x}{y}^{\prime}+\frac{4}{{x}^{2}}y=\frac{x\mathrm{ln}x+1}{x}$

(1) Therefore substituting$x={e}^{t}$ (1) now becomes $y{}^{\u2033}-\frac{3}{{e}^{t}}{y}^{\prime}+\frac{4}{{e}^{2t}}y=\frac{{e}^{\mathtt{+}}1}{{e}^{t}}$ . We need constant coefficients in order to use the variation of parameters method am i right?

Change the variable

It seems a little suspicious that i can factor out 2 and x first.

(1) Therefore substituting

asked 2022-03-16

Assessing stability or instability of a system of equations with complex eigenvalues

Having this system , we get clearly two complex eigenvalues. If one has to assess the stability of the system at these eigenvalues, we have for the matrix A:

$\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$

that ${T}^{2}-4\mathrm{\Delta}\ge {\textstyle \phantom{\rule{1em}{0ex}}}\text{or}{\textstyle \phantom{\rule{1em}{0ex}}}\le {\textstyle \phantom{\rule{1em}{0ex}}}\text{or}{\textstyle \phantom{\rule{1em}{0ex}}}=0$, where $T=a+c$, while $\mathrm{\Delta}=ad-bc=DetA$. But with complex eigenvalues $\pm 2i$, T which must be real, becomes complex and $\delta$ is always 0, when it would vary from greater than or lesser than zero for real eigenvalues. How do we solve this with complex eigenvalues?

asked 2022-04-08

Using a change in variables, Show that the boundary value problem:

$-\frac{d}{dx}\left(p\left(x\right){y}^{\prime}\right)+q\left(x\right)y=f\left(x\right),$

$a\le x\le b,$

$y\left(a\right)=\alpha ,$

$y\left(b\right)=\beta$

Can be transformed to the form:

$-\frac{d}{dw}\left(p\left(w\right){z}^{\prime}\right)+q\left(w\right)z=F\left(w\right),$

$0\le w\le 1,$

$z\left(0\right)=0,$

$z\left(1\right)=0$

Can be transformed to the form:

asked 2022-04-13

Does Lyapunov function need to be defined at zero?

Consider the function$V(x,y)=x-y-y\mathrm{ln}\left(\frac{x}{y}\right)$ , where $x,y>0$ and y is fixed. V is constructed to study the stability of equilibrium point $x=y$ for the system

$\frac{dx\left(t\right)}{dt}=y-x\left(t\right).$

Consider the function

asked 2022-04-01

Solving the ODE

${y}^{\prime 2}-y{y}^{\prime}+{e}^{x}=0$

asked 2022-03-29

How would we solve the same initial value problem but instead of $c\in \mathbb{R}$ being a constant we have a function $f:{\mathbb{R}}^{n}\times [0,t)\to \mathbb{R}$ in its place such that the problem becomes:

$\{\begin{array}{ll}{u}_{t}+b\cdot ({D}_{x}u)+f(x,t)u=h(x,t)& \text{in}{\textstyle \phantom{\rule{0.167em}{0ex}}}{\mathbb{R}}^{n}\times (0,\mathrm{\infty})\\ u(x,0)=g(x)& \text{on}{\textstyle \phantom{\rule{0.167em}{0ex}}}{\mathbb{R}}^{n}\times \{t=0\}\end{array}$

What steps would I take to find a function u(x,t) that satisfies this? I am assuming u(x,t) will be similar to the function I found for the original problem but with some additional integrals (after playing with it for a little), but I am unsure.

I am fairly new to partial differential equations so any help will be appreciated.

What steps would I take to find a function u(x,t) that satisfies this? I am assuming u(x,t) will be similar to the function I found for the original problem but with some additional integrals (after playing with it for a little), but I am unsure.

I am fairly new to partial differential equations so any help will be appreciated.