What is Homogeneous equation: Definition and 34 Discussions
In mathematics, a system of linear equations (or linear system) is a collection of one or more linear equations involving the same set of variables. For example,
3
x
+
2
y
−
z
=
1
2
x
−
2
y
+
4
z
=
−
2
−
x
+
1
2
y
−
z
=
0
{\displaystyle {\begin{alignedat}{7}3x&&\;+\;&&2y&&\;-\;&&z&&\;=\;&&1&\\2x&&\;-\;&&2y&&\;+\;&&4z&&\;=\;&&-2&\\-x&&\;+\;&&{\tfrac {1}{2}}y&&\;-\;&&z&&\;=\;&&0&\end{alignedat}}}
is a system of three equations in the three variables x, y, z. A solution to a linear system is an assignment of values to the variables such that all the equations are simultaneously satisfied. A solution to the system above is given by
x
=
1
y
=
−
2
z
=
−
2
{\displaystyle {\begin{alignedat}{2}x&\,=\,&1\\y&\,=\,&-2\\z&\,=\,&-2\end{alignedat}}}
since it makes all three equations valid. The word "system" indicates that the equations are to be considered collectively, rather than individually.
In mathematics, the theory of linear systems is the basis and a fundamental part of linear algebra, a subject which is used in most parts of modern mathematics. Computational algorithms for finding the solutions are an important part of numerical linear algebra, and play a prominent role in engineering, physics, chemistry, computer science, and economics. A system of non-linear equations can often be approximated by a linear system (see linearization), a helpful technique when making a mathematical model or computer simulation of a relatively complex system.
Very often, the coefficients of the equations are real or complex numbers and the solutions are searched in the same set of numbers, but the theory and the algorithms apply for coefficients and solutions in any field. For solutions in an integral domain like the ring of the integers, or in other algebraic structures, other theories have been developed, see Linear equation over a ring. Integer linear programming is a collection of methods for finding the "best" integer solution (when there are many). Gröbner basis theory provides algorithms when coefficients and unknowns are polynomials. Also tropical geometry is an example of linear algebra in a more exotic structure.
Ignoring the second part of the question for now, since I think it will be more clear once I understand how this equation is homogeneous.
According to my textbook and online resources a first-order ODE is homogeneous when it can be written like so:
$$M(x,y) dx + N(x,y) dy = 0$$
and ##M(x,y)##...
I'm having quite a bit of a problem with this one. I've managed to figure out that ##T_0 = 0##. However, not knowing what ##q(t)## is bothers me, although it seems that I could theoretically solve the problem without knowing it. For ##t>t_1##, integration by parts gives me ##T = Ce^{-t/10}##...
Hi, I'm going to cite a book that I'am reading
Can anyone provide some simple references where I can find at least an intuition regarding what is stated by the author.
Thanks,
Ric
Hi, I have attached part of my steps for solving the homogeneous equation.
The equation is proven to be homogeneous. However after using substitution of y=zx and its' derivative, I was not able to separate the variables conveniently as shown. Please advise. Thank you!
For example, in linear differential equations, there might be these questions where we'd directly use e∫pdx as the integrating factor and then substitute it in this really cliche formula but I never really understood where it came from. Help ?
Hello,
If I have a homgeneous linear differential equation like this one (or any other eq):
$$y''(x)-y'(x)=0$$
And they give me these Dirichlet boundary conditions:
$$y(0)=y(1)=0$$
Can I transform them into a mixed boundary conditions?:
$$y(0)=y'(1)=0$$
I tried solving the equation, derivating...
Homework Statement
Find the general solution of y^{(5)}-y(1)=x
The Attempt at a Solution
I found the complementary function by substitution of the solution form y=e^{kx} giving k=0,1,-1,i,-i, so y_{cf}=a_0+a_1e^x+a_2e^{-x}+a_3e^{ix}+a_4e^{-ix}
Now for the particular integral, the general...
Homework Statement
Show that the homogeneous equation: $$(Ax^2+By^2)dx+(Cxy+Dy^2)dy=0$$ is exact iff 2b=c.
Homework Equations
None, just definitions.
The Attempt at a Solution
Let $$M = Ax^2+By^2$$ and $$N = Cxy+Dy^2$$
Taking the partial derivative of M with respect to y and the partial of...
Homework Statement
Solve y''+(cosx)y=0 with power series (centered at 0)
Homework Equations
y(x) = Σ anxn
The Attempt at a Solution
I would just like for someone to check my work:
I first computed (cosx)y like this:
(cosx)y = (1-x2/2!+x4/4!+ ...)*(a0+a1x+a2x2 +...)...
Suppose ##x_1(t)## and ##x_2(t)## are two linearly independent solutions of the equations:
##x'_1(t) = 3x_1(t) + 2x_2(t)## and ##x'_2(t) = x_1(t) + 2x_2(t)##
where ##x'_1(t)\text{ and }x'_2(t)## denote the first derivative of functions ##x_1(t)## and ##x_2(t)##
respectively with respect to...
Hello.
I forgot the reason why 2nd order differential equation has two independent solutions. (Here, source term is zero) Why 3 or 4 independent solutions are not possible?
Please give me clear answer.
Homework Statement
Solve: A*sin(ωt + Θ) = L*i''(t) + R*i'(t) + (1/C)*i(t). Where: A=2, L = 1, R=4, 1/C = 3 and Θ=45°.
Homework Equations
The system has to be solved by i(t) = ih + ip. I gave the values to A, L, R, 1/C and Θ. I can also give values to ω, but I've come to a doubt when solving...
Homework Statement
The system is declared as follows:
8/(2*x - y) - 7/(x + 2*y) = 1
4/((2*x - y)^2) - 7/((x + 2*y)^2) = 3/28
Homework Equations
The Attempt at a Solution
I define 'x' to equal k*y and I replace it inside the equation:
8/(2*k*y^2) - 7(k*y + 2*y) = 1...
I need help finding a linear homogenous constant-coefficient differential equation with the given general solution.
y(x)=C1e^x+(C2+C3x+C4x^2)e-x
2. I tried to come with differential equation but this is it
I can 't seem how to begin
Actually I can't find if a differential equation is homogeneous or not
I thought homogeneous is given by
dy/dx= f(x,y)/ g(x,y)
but it doesn't look like that
For eg:
dy/dx= (y+x-1)/(y-x+2) is not homogeneous at all though
f(x,y)=y+x-1 and g(x,y)=y-x+2
How can you tell...
My doubt is that is dimension of a 2nd order homogeneous equation of form y''+p(x)y'+q(x)=0 always 2 ? or dimension is 2 only when p(x),q(x) are contionuos on a given interval I..??
Hi, I don't understand why the general solution of 2nd order homogeneous equation is linear? Why is c_1e^(xt)+c_2e^(xt) a linear differential equation? What am I missing here? Any help would be appreciated, I'm struggling a bit understanding the concepts of differential equations...
Homework Statement
xdx+sin\frac{y}{x}(ydx-xdy) = 0
The Attempt at a Solution
Well, it's quite easy. But I'm quite confused if this is homogeneous or not, because of the sine function. This is my solution, assuming that this is a homogeneous equation.
let x = vy, dx = vdy + ydv; then...
Homework Statement
The problem is setting up the equation, it says that the matrix equation will be made up of four equations for the 2 unknowns.
I'm supposed to find for which a's and b's the equation is true, using a linear system and gaussian elimination.
Homework Equations
A2 + aA + bI2 =...
i am having trouble finding the general solution for the given homogeneous equation:
x2yy' = (2y2 - x2)
which i made into
x2dy = (2y2 - x2) dx
i turned it into the following:
(2y2 - x2) dx - x2 dy = 0
then i used substitution of y = xv and got
(2(xv)2 - x2 - x2v) dx - x3 dv = 0
then...
Homework Statement
The problem has to do with diagonalizing a square matrix, but the part I'm stuck on is this:
Bx=0, where B is the matrix with rows [000], [0,-4, 0], and [-3, 0, -4].
After performing rref on the augmented matrix Bl0, I get rows [1,0,4/3,0], [0,1,0,0], [0000].
I am...
Homework Statement
y'' + y = -2 Sinx
Homework Equations
The Attempt at a Solution
finding the homogeneous solution, is simple;
yh(x) = C1 Cos(x) + C2 Sin(x)
for the particular solution,
I let y = A Cos(x) + B Sin(x)
thus, y' = -A Sin(x) + B Cos(x)
y'' = -A Cos(x) - B...
Homework Statement
Find y as a function of x if
y'''−11y''+28y'=0 y(0)=1 y'(0)=7 y''(0)=2
I have one attempt left on this question. Could someone verify my answer for me?
Homework Equations
The Attempt at a Solution
(use t as lamda)
t^3-11t^2+28t=0...
Homework Statement
(4y4-9x2y2-144)dx - (5xy3)dy = 0
Homework Equations
substitute y = xv
dy = dx v + dv x
The Attempt at a Solution
after substituting i got
(4x4v4-9v2x4-14x4)dx - (5v3x4)dx.v + dv.x
= (4v4-9v2-14)dx - 5v3(dx.v + dv.x) = 0
= dx(4v4-9v2-14-5v4)+dv(-5v3x)= 0...
Homework Statement
y2dx -x(2x+3y)dy =0 I have to recognize the equation and solve it
Homework Equations
The Attempt at a Solution
I did y2dx - (2x2+ 3yx) dy=0
which is a homogeneous now
after I substitude x=uy
dx=udy + ydu
I stuck here after the substitution...
First Order Linear Non-Homogeneous Equation
Homework Statement
I need to solve for e(t)
Homework Equations
Do I use Laplace Transform for the last integral?
The Attempt at a Solution
\begin{subequations}
\begin{eqnarray}
\nonumber
\dot{\hat{{\cal E}}}(t) &=& -\kappa...
this is the given equation
y'=(4y-3x)/(2x-y)
and here is all the work I've done so far:
(4v-3)/(2-v)=v+x*dv/dx
i moved v over and came up with this
(-3+2v+v^2)/(2-v)=x*dv/dx
did a flip
(2-v)/(-3+2v+v^2)dv=dx/x
by partial fractions I got a=-3/2 and b=1/2
so...
Homework Statement
Given,
(y+2)dx + y(x+4)dy = 0, y(-3) = -1Homework Equations
v=y/xThe Attempt at a Solution
I've been REALLY struggling with homogeneous equations for some reason...I just don't understand them all.
so far I've tried two things.
(1)dx -(y)dy
----- -------
(x+4)...
y''(t)+A^2y(t)=f(t), t>0, y(0)=B, y'(0)=C, A, B, C\in\mathbb{R}
e^{iAt} is a particular solution of the homogeneous equation. I can multiply it by some arbitrary function and find another solution of the homogeneous case, but when I try with the f(t) on the RHS, I can't do it. Anyone help?
Hey...
So the question is as stated:
Show that
\frac{1} {M_x + N_y} , where M_x+N_y is not identically zero, is an integrating factor of the homogeneous equation M(x, y)dx+N(x, y)dy=0 of degree n.
So I am not too sure where to go with this. I suppose what it's saying is, that I'm...
(x^2 + y^2)dx + (2xy)dy = 0
I get y = sqrt((kx^5 + x^2)/3) Where k = c2 cubed, and c2 = ln(c) so k = 3ln(c)
But, the answer the teacher gave is (x^2)(y^3) - x - ln(y) = c I can't come up with anything remotely close. I know this isn't in a pretty LaTeX form, but I am new and haven't...
Can someone explain what the homogeneous equation is :redface: and how do you find the 'null vectors' and hence the general solution.
Eg.
AX =
[6]
[8]
[4]
A =
[1 2 4]
[3 1 2]
[0 2 4]
X =
[2]
[0]
[1]
Find the null vectors of A and general solution.