How do I solve a non-homogeneous Laplace equation?
Non Homogeneous 2nd order Partial Differential Equation is named Poisson's equationHere f(x,y,z) is a function of source( e.g. point mass, point charge).Green's function is the inverse of a differential operator (in a more general often necessary for theoretical physics and proper mathematical context, this is a distribution). So, the differential operators are mostly the Laplace, the Poisson, and the d'Alembert operator with given boundary condition.I will solve Poisson Eqn by the Green's function method.Since the source term is at a point through the whole volume with surface then we definewhere δ^3 is 3D Dirac delta function . So ,&From Green's identities using Green's 2nd integral theorem,Construction of Green’s Function :Method 1 : (Gauss Divergence Theorem)Thus,Method 2 : ( Using Fourier Transform)A convenient and efficient way to construct Green's functions of such operator is to use Fourier transform which can be applied for a wider class of differential operators .From Definition of Fourier Transform,Now.= >So,Therefore, Solution of Poisson Eqn,Another Ways To Solve Poisson Eqn :For numerical solve :Samim Ul Islam's answer to How can you use finite difference approximation to solve Poisson's equation?For more :Green's function - WikipediaGreen's Function--Poisson's EquationSolving Poisson's equation with the help of Greens functionhttp://www.dfcd.net/articles/gre...https://www.ndsu.edu/pubweb/~nov...http://www.srl.caltech.edu/phys1...https://ocw.mit.edu/courses/math...https://web.stanford.edu/class/m...http://www.damtp.cam.ac.uk/user/...https://www2.ph.ed.ac.uk/~dmaren...
Nonhomogeneous linear equation help?
Consider the nonhomogeneous linear equation of the form dy/dt+a(t)y=b1(t)+b2(t) that is, b(t) is written as the sum of two functions. Suppose that yh(t) is a solution of the associated homogeneous equation dy/dt+a(t)y=0, that y1(t) is a solution of the equation dy/dt+a(t)y=b1(t), and that y2(t) is a solution of the equation dy/dt+a(t)y=b2(t). Show that yh(t)+y1(t)+y2(t) is a solution of the original nonhomogeneous equation. Please help, I'm SO lost, I have no clue where to start... I know that when you add the natural response and forced response (yh(t) and y1(t)+y2(t), respectively) you get a solution, but I only know this because my professor said it, I don't know how to show it mathematically.
What are homogenous and non-homogeneous equations?
I want to make this answer very small and simpleEquations which have both the reactants and products of the state of matter( say gases) are called homogenous equations others are heterogeneous…..P.S: I hope you mean chemical equations and not mathematical equations———CGPS———
Is y'=cos2x homogeneous or a non-homogeneous equation, and why?
As a linear differential equation of first order, it is non-homogeneous, because there is a term cos(2x), which does not involve y or y'.
Differential Equations (Non - Homogeneous) HELP!?
Solve the initial value problem: y" - y' + 2y = x+ 1 y(0)=3 ; y'(0) = 0 if you have time can you help me with this one too? Find a solution; y" + 2y' - 3y = 1 + xe^x Much love if you can help me!! thank you
Find the general solution of the given nonhomogeneous equation.?
x²y'' + xy' + (x² - ¼)y = √x³ y₁ = ( cos(x) / √x ) ==> y₁' = -½ ( 2xsinx + cosx ) / √x³ y₂ = ( sin(x) / √x ) ==> y₂' = ½ ( 2xcosx - sinx ) / √x³ Wronskian: W(y₁ ,y₂ ) = (cos(x) / √x )( ½ ( 2xcosx - sinx ) / √x³ ) + (sin(x) / √x )(½ ( 2xsinx + cosx ) / √x³) W(y₁ ,y₂ ) = 1/x Y = -y₁ ∫ (y₂ √x³ )/ W(y₁ ,y₂ ) dx + y₂ ∫ (y₁ √x³ )/ W(y₁ ,y₂ ) dx Y = -y₁ ∫ sin(x) dx + y₂ ∫ cos(x) dx Y = y₁cos(x) + y₂ sin(x) Y = 1/√x general soln y(x) = c₁cos(x) / √x + c₂sin(x) / √x + 1/√x y(x) = ( c₁cosx + c₂sinx + 1) / √x
Find the general solution of the given nonhomogeneous equation?
The general solution can be found by variation of parameters. This method assumes that for a 2nd order nonhomogeneous equation the solution has the form: y(x) = u₁(x)∙y₁(x) + u₂(x)∙y₂(x) where y₁ and y₂ are a two linearly independent solution of the associated 2nd order homogeneous equation. The functions u₁ and u₂ are given by the integrals: u₁(x) = - ∫ f(x)∙y₂(x)/W(x) dx u₂(x) = ∫ f(x)∙y₁(x)/W(x) dx f is the function on right hand side of the differential equation, i.e. the function of x which makes the equation nonhomogeneous. W is the Wronskian of y₁ and y₂. For details see [1] f(x) = x^(3/2) y₁(x) = x^(-1/2)∙cos(x) y₂(x) = x^(-1/2)∙sin(x) => y₁'(x) = - (1/2)∙x^(-3/2)∙cos(x) - x^(-1/2)∙sin(x) y₂(x) = - (1/2)∙x^(-3/2)∙sin(x) + x^(-1/2)∙cos(x) => W(x) = y₁(x)∙y₂(x) - y₁'(x)∙y₂(x) = ( x^(-1/2)∙cos(x) )∙( - (1/2)∙x^(-3/2)∙sin(x) + x^(-1/2)∙cos(x) ) - (- (1/2)∙x^(-3/2)∙cos(x) - x^(-1/2)∙sin(x) )∙(x^(-1/2)∙sin(x)) = ( - (1/2)∙x^(-2)∙sin(x)∙cos(x) + x^(-1)∙cos²(x) + (1/2)∙x^(-2)∙sin(x)∙cos(x) + x^(-1)∙sin²(x) = x^(-1)∙(sin²(x) + cos²(x)) = x^(-1) = 1/x => u₁(x) = - ∫ x^(3/2) ∙ x^(-1/2)∙sin(x) / (1/x) dx = - ∫ x^(2) ∙ sin(x) dx Integrating by parts twice gives: u₁(x) = -2∙x∙sin(x) + (x^(2) - 2)∙cos(x) + C₁ u₂(x) = ∫ x^(3/2) ∙ x^(-1/2)∙cos(x) / (1/x) dx = ∫ x^(2) ∙ cos(x) dx Integrating by parts twice gives: u₂(x) = (x^2 - 2)∙sin(x) + 2∙x∙cos(x) + C₂ => y(x) = u₁(x)∙y₁(x) + u₂(x)∙y₂(x) = ( -2∙x∙sin(x) + (x^(2) - 2)∙cos(x) + C₁ )∙( x^(-1/2)∙cos(x) ) + ( (x^2 - 2)∙sin(x) + 2∙x∙cos(x) + C₂ )∙( x^(-1/2)∙sin(x) ) = C₁∙x^(-1/2)∙cos(x) - 2∙x^(1/2)∙sin(x)∙cos(x) + (x^(2) - 2)∙x^(-1/2)∙cos²(x) + C₂∙x^(-1/2)∙sin(x) + (x^2 - 2)∙x^(-1/2)∙sin²(x) + 2∙x^(1/2)∙sin(x)∙cos(x) = C₁∙x^(-1/2)∙cos(x) + C₂∙x^(-1/2)∙sin(x) + (x^(2) - 2)∙x^(-1/2)∙(cos²(x) + sin²(x) ) = C₁∙x^(-1/2)∙cos(x) + C₂∙x^(-1/2)∙sin(x) + (x^(2) - 2)∙x^(-1/2)
2nd order nonhomogeneous DE?
y'' - 7y' + 10y = 24e^(x) y = c₁e^(2x) + c₂e^(5x) + 6e^(x) i) differentiate y (wrt x) y = c₁e^(2x) + c₂e^(5x) + 6e^(x) y' = 2c₁e^(2x) + 5c₂e^(5x) + 6e^(x) ii) find the second derivative of y y = c₁e^(2x) + c₂e^(5x) + 6e^(x) y' = 2c₁e^(2x) + 5c₂e^(5x) + 6e^(x) y'' = 4c₁e^(2x) + 25c₂e^(5x) + 6e^(x) substitute the differentiated ys into the original differential equation y'' - 7y' + 10y = 24e^(x) y'' - 7y' + 10y = 24e^(x) [ 4c₁e^(2x) + 25c₂e^(5x) + 6e^(x) ] - 7[2c₁e^(2x) + 5c₂e^(5x) + 6e^(x) ] + 10 [c₁e^(2x) + c₂e^(5x) + 6e^(x) = { 4c₁e^(2x) - 14c₁e^(2x) + 10c₁e^(2x) }+ { 25c₂e^(5x) - 35c₂e^(5x) + 10c₂e^(5x) } + { 6e^(x) - 42e^(x) + 60e^(x) } = 0 + 0 + 24e^(x) = 24e^(x) After differentiating and replacing the values in the original differential equation the left-hand equates to the right-hand side; thus y=c1e^2x+c2e^5x+6e^x on (-oo, oo) is the solution to the differential equation y"-7y'+10y=24e^x