how to use linear algebra to solve integration by parts problems?
463328
Your question is not clear. Furnish some examples.
1069671
yep...scalars multiply vectors component by component just like you have shown.  you can also show it by plotting out the original vectors, scaling them appropriately, adding them, and finding the resukltant components.
677393
it is probably easiest to expand into a polynomial like bboy does. or use u substitution:\nlet u =1-3x\ndu=-3 dx\nso integral of -2/3 u^2 = -2/9u^3 = -2/9(1-3x)^3
171039
integrate it by parts : \nint[U.dv] = [U.V] - int[V.du]\nwhere U = x, dv = sin^-1
115853
#include <iostream>\n#include <vector>\nusing namespace std;\n#define rep(i, n) for((i) = 0; (i) < (n); (i)++)\n#define rep2(i, k, n) for((i) = k; (i) < (n); (i)++)\nvoid f21();\nint n;\ndouble mat[100][100];\nint main()\n{\nwhile(true)\n{\ncin >> n;\nif(!n) break;\nmemset(mat, 0, sizeof(mat));\nf21();\n}\nreturn 0;\n}\n\nvoid f21()\n{\ncout << "****************************************" << endl;\ncout << "*          INVERSE OF MATRIX           *" << endl;\ncout << "****************************************" << endl;\n\ncout << "ORDER OF THE MATRIX: " << n << endl;\ncout << "ENTER ELEMENT ROW-WISE" << endl;\nint m = 2*n;\nint i, j, k;\nrep(i, n) {\nrep(j, n) {\ncin >> mat[i][j];\ncout <<"&#92;t";\nprintf("%5.3lf", mat[i][j]);\n}\nrep2(j, n, m) {\nif(i == (j-n)){\nmat[i][j] = 1;\ncontinue;\n}\nmat[i][j] = 0;\n}\ncout << endl;\n}\n\ncout << endl << "INVERSE MATRIX: " << endl;\ndouble pivot, comm;\nrep(i, n) {\npivot = mat[i][i];\nif(mat[i][i] == 0){\ncout << "DIVISION BY ZERO." <<endl;\nreturn;\n}\nrep(j, m) mat[i][j] = mat[i][j]/pivot;\nrep(j, n) {\nif(i == j) continue;\ncomm = mat[j][i];\nrep2(k, i, m) {\nmat[j][k] = mat[j][k] - (comm*mat[i][k]);\n}\n}\n}\n\nrep(i, n) {\nrep2(j, n, m) {\ncout <<"&#92;t";\nprintf("%6.4lf", mat[i][j]);\n}\ncout << endl;\n}\ncout << endl;\n}\n\n/*This is writen in C++ this will output the inverse of a Matrix\ninput:\n4\n1 2 3 4\n2 1 4 2\n2 5 7 12\n1 -1 3 11\n\n5\n1.173 -0.634 0 0 0\n-0.539 1.173 -0.634 0 0\n0 -0.539 1.173 -0.634 0\n0 0 -0.539 1.173 -0.634\n0 0 0 -0.539 1.173\n\n0\n*/
403929
Integration by parts is done by breaking down an integral so that\n\nS u dv = uv - S v du (where S is the integration symbol)\n\nFor example, if we were to integrate xcos x dx,\n\nlet u = x\ndu = dx\n\nthen dv = cos x (all the other stuff we didn't use for u)\nv = sin x (just integrate dv)\n\nand the formula will look like\n\nS xcos x dx = xsin x - S sin x dx\n= xsin x + cos x + C\n\nIt's kinda an art form or experience in picking which part of the integral to make u and dv - there's really no rule that says u = cos x.  Experience (or trial and error) will show you that that's probably not the better choice though .
274355
Depends on the system you are trying to solve. check your limits of integration. from -infinity to 0 plus 0 to infinity.
210912
Put the left side coefficients into matrix [A],\n[6 2]\n[5 6]\nand the right side values into matrix [B]\n[-16]\n[..4]\nthen tell the calculator to multiply [A] inverse * [B]\nthe result is the solution matrix\n[x]\n[y]\n\nnote, the inverse key looks like x^-1.  \n\nsolve the second system the same way.  coefficient matrix will be 3x3, answer matrix 3x1.
787521
Example:\nIf z=f(u,v),\nu=g(x,y) and\nv=h(x,y), then\n(all derivatives are supposed to be partial derivatives)\ndz/dx=dz/du*du/dx+dz/dv*dv/dx\ndz/dy=dz/du*du/dy+dz/dv*dv/dy\n\nAlternatively, you can set of the matrices\n[[dz/du dz/dv]]\nand\n[[ du/dx du/dy]\n[dv/dx dv/dy]]\nand get that\n[[dz/dx dz/dy]]\nis the product of the two matrices.
145305
Use cubic equation \ny=ax3+bx2+cx+d\n\nSince you have four points data you can get four linear equations with four variables which is very easy to solve, get variables (a,b,c,d) and plug the values to cross check and see u will get answer. Use guass jordan or elimination method.
1068054
well Y' = dy/dx\nso using seperation of veriables we can get\ndy/dx = K - 1/3*y^3\nand dy/(K-1/3*y^3)=dx now integrate both sides. \nDoes that help?
158690
substitute x=e^y\nthen use a table of integration formulas?
985272
Solving by parts means you see your integral as a product of two functions and you integrate one function and derivate the other. Usually, when you have ln(t) in your integral, you know it's the one you have to derivate. \nSo you need to find F=integral (2(t^2) * ln(t) dt) = 2* integral (v' *u dt) with:\nu = ln t ................                                                u'=1/t\nv' = t^2 ................                                                v= (t^3)/3\nSince we're doing an integration by parts we have:\nF= 2* [ln t * (t^3)/3] - 2* integral (1/t * (t^3)/3) (the 2 factor is due to your initial integral and you have to put the boundaries on every integral too)\nI found 1/3 but I'm not sure. If you still have questions go ahead and ask\nHope this helps a bit
719380
Let y = x^(1/2)\n\nso y^2 = x,\n\ndx = 2y dy\n\nThe problem becomes:\n\nint [1 / (y + y^3)] 2y dy\n\nOr:\n\nint 2 [1 / (1 + y^2)] dy\n\nA recognizable integral. Integrate and substitute back x^1/2 = y
591447
Well ... thought I had solved it, but then I got stuck!!!  Integration is not the best thing I am at!!!!\n\nUsing "$" as the integral sine.\n\nUse integration by parts: $udv = uv - $vdu\n\nThen split the orignal question into:\n\ny = $ (2x-3)(1/(√(9-x^2))) dx\n\nu = 2x - 3\ndu = 2\n\ndv = 1 / √(9 - x^2)\ndv = 1 / √(3^2 - x^2)          ==> standard integral\nv = sin^-1(x/3)     \n\nThen use the integration by parts equation above to try and solve it!  \n\ny = (2x-3)(sin^-1(x/3)) - $2 / sin(x/3)\n\nThen you have to solve that integral by parts again (using "z" this time, so not to be confused with the whole equation which is equal to "y")\n\nz = $2 / sin(x/3)\n\nu = 2\ndu = 0\n\ndv = 1 / sin(x/3)\nv = 3[x / 3sin(x/3) + √(1 - (x/3)^2)]\n\nz = $2 / sin(x/3) = 6[x / 3sin(x/3) + √(1 - (x/3)^2)]\n\nOK, (I hope this is right) so ... then put that back into the above equation that I got after the first round of integrating by parts!\n\nThat is: y = (2x-3)(sin^-1(x/3)) - $2 / sin(x/3)\n\ny = (2x-3)(sin^-1(x/3)) - 6[x / 3sin(x/3) + √(1 - (x/3)^2)]\n\nWhich eventually simplifies down to:\n\ny = [(4x - 3)/(sin(x/3)] - 6√(1 - (x/3)^2)]\nor\ny = [(4x - 3)/(sin(x/3)] - √(36 - 4x^2)]\n\nLet me know if that was right, if you ever find out!!!!\n\nI hope it was ... good luck
897617
This is an equation with separated variables. Rewrite as \n\ndu/u^3 = (t^3-t)dt \n\nand integrate (find antideritive) of each side separately.
815728
first this is not a PDE but ODE\n\nsimple integration would do