Find two linearly independent solutions of 2x^2y''-xy'+(1-6x)y=0 of the form?
Find two linearly independent solutions of 2x^2y''-xy'+(1-6x)y=0 of the form where y1=x^(r1)(1+a1x+a2x^2+a3x^3...), y2=x^(r2)(1+b1x+b2x^2+b3x^3...) r1>r2 Find r1=,a1=,a2=,a3= r2=,b1=,b2=,b3=
Is the vector space {sin(x), sin(2x), sin(3x)} linearly independent or linearly dependent?
set a linear combination of your 3 vectors =0 and prove that the numbers have to be =0: A sin(x) + B sin(2x) + C sin(3x) = 0 i.e. A,B,C are numbers such that the linear combination is the constant zero. now take x=pi/2, then 0= A sin(pi/2) + B sin(2pi/2) + C sin(3pi/2) = A - C, so A=C. Now x=pi/4 0= A sin(pi/4) + B sin(2pi/4) + A sin(3pi/4) = A/sqrt(2) + B + A/sqrt(2) = 2 A/sqrt(2) +B = sqrt(2) A + B, so B= -sqrt(2) A Now, x = 3pi/4 0= A sin(3pi/4) + B sin(2(3)pi/4) + A sin(9pi/4) = A/sqrt(2) - B + A/sqrt(2) =sqrt(2) A - B, but then B = sqrt(2) A which means that -sqrt(2) A = sqrt(2) A so A =0, therefore B=0 and C=0 '
Three vectors that lie on the plane are linearly dependent. How do I prove this statement?
Since Prashant already answered your question I'll try to answer it in a cool way. Let your three vectors be A, B and C.Recall the scalar triple product:Since A, B and C are coplanar, A●(BxC) = 0. Thus, the matrix shown above with rows equal to the components of the three vectors has determinant 0 and is thus singular. Singular matrices have incomplete rank, thus the vector set formed by the row elements is linearly dependent.
Linearly Independent generators?
From x₁ - x₂ - x₃ = 0, we obtain x₁ = x₂ + x₃. Treat x₁ as a basic variable that's expressed in terms of the free variables x₂ and x₃. Now let x be an arbitrary vector in the given subspace. Then we have: x = (x₁, x₂, x₃) = (x₂ + x₃, x₂, x₃) = (x₂, x₂, 0) + (x₃, 0, x₃) = x₂(1, 1, 0) + x₃(1, 0, 1) Thus, any arbitrary vector x can be expressed as a linear combination of the vectors (1, 1, 0) and (1, 0, 1). Note that these two vectors are linearly independent, since neither vector is a scalar multiple of the other. Thus, one possible set of linearly independent generators would be: {(1, 1, 0), (1, 0, 1)}.
Help finding the linearly independent functions that are annihilated ...?
The annihilator of a linear combination of functions is the product of the annihilators of the individual functions. Here we have the product of three annihilators, D^2, (D-5), and (D-7). D^n is the annihilator of a polynomial of degree n+1, so D^2 annihilates the function w(x) = a*x^3 + b*x^2 + c*x + d, where a, b, c, and d are constants. (D - a)^m is the annihilator of the function exp(ax)*(polynomial of degree m), so (D-5) is the annihilator of y(x) = f*x*exp(5x) and (D - 7) is the annihilator of z(x) = g*x*exp(7x). The three linearly independent functions are w(x), y(x), and z(x) given above. The annihilator D^2 * (D - 5)(D - 7) zeros the function: w(x) + y(x) + z(x) = a*x^3 + b*x^2 + c*x + d + f*x*exp(5x) + g*x*exp(7x)
Determind whether three vectors are linearly independent.?
linear independence means that for the vectors A, B, and C, there are no constants a, b or c other than 0 which satisfy aA + bB = cC. In this case, we can see that 2A + B = C, so they are not linearly independent A way of checking if they are is to take a matrix with the columns being the vectors as so (note: you could also make the vectors the rows, it doesn´t matter, I just prefer to work in columns): 1 1 3 0 2 2 1 3 5 compute its determinant. in this case the determinant = 10 + 2 - (6 + 6) = 0. If the det is 0, then they are lin dep, if it is non 0, they are lin indep