Calculus 3 and Differential Equations?
146580
I would say to take calc 3 only.  Taking both classes may be overwelming and you may end up dropping.  Also, like people have said before, calc 3 is similar to calc 1 but in higher dimesion so a lot of the concepts are similar.  Diff eq is a class that you might want to take over a whole semester since there are more new concepts to digest.
421380
from first equation put the value of b in second and then form a second degree differential equation in "a"...good luck!
757602
[(x^3)/(2 - x)]^4/3 \ndy/dx = (4/3)[(x^3)/(2 - x)]^1/3[3x^2(2 - x) + (x^3)]/(2-x)^2\ndy/dx = (4)[(x^3)/(2 - x)]^1/3[6x^2 - 3x^3 + (x^3)]/(3(2-x)^2)\ndy/dx = 8x^3(3 - x)/(3(2-x)^7/3)
196268
(x^2+1)dy/dx+2xy=4x^2 \n---> (x^2+1)dy+(2xy-4x^2)dx=0\n\nIt is a perfect differential because:\n\n∂(x^2+1)/∂y=∂(2xy-4x^2)/∂x\n\n\n--->d(yx^2+y-4x^3/3)=0\n\n--->yx^2+y-4x^3/3=c --->\n y= (c+4x^3/3)/(1+x^2)\n\nNow we check answer:\n\ndy/dx= \n[4x^2*(1+x^2)-2x*(c+4x^3/3)]/(1+x^2)^2\n\n---> (x^2+1)dy/dx+2xy = \n\n[4x^2*(1+x^2)-2x*(c+4x^3/3)]/\n(1+x^2)+2x(c+4x^3/3)/(1+x^2)\n\n=[4x^2*(1+x^2)-2x*(c+4x^3/3)+2x(c+4x^3/3)]/\n(1+x^2)\n\n=4x^2   so it is the correct answer.\n\nSince ordinary homogeneous differential equations can\nhave only one general solution, our answer is the one.
197729
Are you sure you wrote the problem right?  It is at y=3 and not x=3?\n\nA previous poster solved for y(x) when x=3 and got 38 (which is correct) but that's not what you asked.\n\nI think you will have to solve x^3 + x^2 = 3 to get a value for x before you can go on.\n\nAfter that, it's pretty straight forward.  You would find y'(x) to get your slope.  Then the equation of the tangent line of f(x) at x=a is y=f(a)+m(x-a).
274505
dy/du=6u^2\ndy/du=(2x^2)/3
71733
Take y on the other side\n\ny^2 dy/dx + y^3 = e^(x)\nMultiply by 3\n\n3 y^2 dy/dx + 3 y^3 = 3 e^(x)\n\nFirst term is d/dx (y^3)\nHence put z = y^3 and teh equation becomes\ndz/dx + 3z = 3 e^(x)\n\nNow this is in standard form\nMultiply throughout by e^(3x)\n\ne^(3x) dz/dx+ 3 e^(3x) z = 3 e^(4x)\nLHS can be re-written as\n\nd/dx { e^(3x)* z) = 3 e^(4x)\nIntegrate both sides\ne(3x)*z = 3 e^(4x)/4 +C\n\nz = 3/4 e^(4x)*e^(-3x) + C e^(-3x)\n\nSubstitute back for z=y^3
714362
Let u be 3e^x + 1, then du/dx = 3e^x .  So your integral amounts to \n\nintegral( (1/3 du/dx) u^(1/3)) = 1/3 integral(u^(1/3)du)\n\n= (1/3)*1/(1+(1/3))*u^(1 + 1/3) + C\n=(1/3)*(1/(4/3))*u^(4/3) + C\n= (1/3) * (3/4) * (3e^x+1)^(4/3) + C\n= (1/4)* (3e^x + 1)^(4/3) + C\n\nCheck by  differentiating:\n\nd/dx (1/4 (3e^x + 1)^(4/3) + C) = \n(4/3)* (1/4) * (3e^x + 1)^(1/3) * (d/dx (3e^x + 1)) + d/dx(C) =\n(4/12) * (3e^x + 1)^ (1/3) * 3e^x dx + 0 + 0 =\n(1/3) * (3e^x + 1)^(1/3) * 3e^x dx =\ne^x (3e^x + 1) dx\n\nwhich is what you started with.
886519
The first one is non-linear (because of the (y')^2 term) and I am not an expert on non-linear differential equations.  Are you sure that is right because if it is only y', then you can simply seperate variables and then solve it.\n\nThe second one, find the integrating factor and then it becomes an exact equation.
12581
A first-order differential equation of the form:\n\na1(x)*(dy/dx) + a0(x)*y = g(x)\n\nis a linear equation.\n\nWhen g(x) = 0, it's homogenous.\n\nSo, (x^2-9)(dy/dx) + xy = 3 is linear and similarly, (x^2-9)(dy/dx) + xy = 0 is homogenous.\n\nA first-order differntial equation of the form\n\ndy/dx = g(x)h(y)\n\nis separable.\n\nSo dy/dx = (y^2)x(e^(3x+4y)) is seperable since (y^2)x(e^(3x+4y)) = (x(e^3x))((y^2)(e^4y)), and dy/dx = y + sin(x) is NOT separble since it cannot be defined as a product of two equations seperately of x and y.\n\nI can't recall how to discern if a diff equation is exact, but this webpage\n\nhttp://www.utpb.edu/scimath/wkfield/mod2/Exact.html\n\ngives a very detailed explanation.\n\nHope that helps.
823232
y = (3x - 2)/(5x² + 4)\n\nW.w.t.f dy/dx.  You can use the formula:\nd/dx (u/v) = (v du/dx - u dv/dx)/v²\n\nNow, we have u = 3x - 2, v = 5x² + 4.  du/dx = 3 and dv/dx = 10x. We can now substitute them to the formula:\ndy/dx = [(5x² + 4)(3) - (3x - 2)(10x)]/(5x² + 4)²\n\nNow,\ndy/dx = (15x² + 12 - 30x² + 20x)/(5x² + 4)²\n\nThus,\ndy/dx = (-15x² + 20x + 12)/(5x² + 4)²\n\ntherefore,\ndy/dx = -(15x² - 20x - 12)/(5x² + 4)²\n\n^_^
806346
using implicit differentiation:\nyou get 2y(dy/dx) - [y + x(dy/dx)] -3 = 0\nthen move 3 over and factor out (dy/dx):\n(2y-x)(dy/dx)=3 + y\ndy/dx=(3+y) / (2y-x)\nnow plug in your point (0,-1)\ndy/dx = (3+ -1)/(2(-1) - 0\ndy/dx = -1\nanswer = -1\n\n(dont forget to use your chain rule when you are implicitlly differentiating)\n\n#2\nsame thing make sure you sue multiplication rule\n[y^2 + 2xy(dx/dy)] - 2(dx/dy) + 12y^2(dx/dy) = 0\nthen move over the y^2 and factor out dx/dy\n(dx/dy)[2xy - 2 + 12y^2] = y^2\ndx/dy = y^2/ (2xy -2 +12y^2)\nPlug in 1 for y\ndx/dy = 1/(2x -2 + 12)\ndx/dy = 1/ (2x +10) or (2x + 10)^(-1)\n\nAnswer (leave in terms of x) = 1/ (2x +10) or (2x + 10)^(-1)\n\nI may have made mistakes so let me know
1139073
Differential Equations
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?
134530
First, generate 3 equations by inserting the values of x and y for each of the coordinates.\n\nEqn 1. For (-1,1): 1=a*-1^2+b*-1+c yields: 1=a-b+c\nEqn 2. For (1,3): 3=a*1^2+b*1+c yields: 3=a+b+c\nEqn 3. For (3,2): 2=a*3^2+b*3+c yields: 2=9a+3b+c\n\nStep 1: solve Eqn 1. for a:  a=1+b-c\nStep 2: insert a from Step 1 into Eqn 2.: 3=(1+b-c)+b+c ==> 2=2b ==> b=1\nStep 3: insert b=1 back into Step 1 and re-solve for a: a=1+1-c ==> a=2-c\nStep 4: insert the results from Step 2 for b and Step 3 for a into Eqn 3 to find c.\n  2=9*(2-c)+3*1+c ==> 2=18-9c+3+c ==> 19=8c ==> c=19/8\nStep 5. Insert the results from Step 4 and Step 2 into Eqn 1 to find a.\n  1=a-1+19/8 ==> a=-3/8\n\nTherefore:\n  a=-3/8\n  b=1\n  c=19/8\n\nStep 6: Verify for -3/8*x^2 + 1*x +19/8\n (-1,1): 1=-3/8*(-1^2)+1*-1+19/8\n (1,3): 3=-3/8*(1^2)+1*1+19/8\n (3,2): 2=-3/8*(3^2)+1*3+19/8
184556
Let x(t) = A*cos( w*t + theta )\n\nthen x(0) = A*cos( theta )\nand dx/dt (0) = -A*w*sin(theta)\n\nso if you know x and its derivative in t=0 u have two equations and two unknowns ( A and theta ) . the rest is simple algebra .
701308
no.\nif u=tan3x you take the derivative of the function:\n(sec 3x)^2 and then multiply it by the derivative of what's INSIDE the function (the chain rule). THe derivative of 3x is 3.\ndu= 3(sec 3x)^2.\n\nHere's the chain rule explanation\nif u=f(g(x)) then du= f'(g(x)) * g'(x)\nin words:\ndu is f prime of the original function multiplied by the deriv of what's inside the fuction.