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# 4c

We can find the vector equation of a three dimensional plane given either three points in the plane, two vectors and a point on the plane, a point on the plane and two points or the equation of the plane in some formulation. The vector equation of a plane is of the form c=[]=[] (0 ss)r0(arrow overhead)+μv(arrow overhead) = λw(overhead). By some means obtain both v(arrow overhead) and w(arrow overhead). If you have three points in the plane A,B and C then you can take r0 to be any of (arrows overhead)OA,OB or OC with (arrow overhead)v =
Partial fractions: a question may seek for the the expression of fractions such as 1/x²-2x-8 in the form A/Cx+F + B/Ex+F [these are partial fractions); as x²+2x-8 factorises in (x+4)(x-2): 1/x²+2x-8 = 1/(x+4)(x-2) = A/ x+4 + B/x-2; to get 1=A(x+o)+B(x+p), multiply through by (x+4)(x-2) to get 1=A(x-2)+B(x+4) after cancellation. Eliminate A by having x equal to 2 to get b= 1/6, then eliminate B by having x equal to -4 to get A= -0.5. Thus 1/x²+2x-8 = -1/2(x+4) + 1/6(x-2).
If we have a denominator which includes an quadratic term which does not factorise, include, in the answer a term Dx+E/Ax²+Bx+C. In general we have one term for every term in the denominator which does not factorise, and for that term, the highest exponent of x in the numerator is less than the highest exponent of x in the denominator by 1. When trying to find the value of variables, variables of which the value is known may be replaced by their value.
To put 5(x + 2)/(x+1)(x+6) into partial fractions using the cover up method: cover up the x + 6 and substitute, into what's left; -6, giving 5(-6 + 2)/(-6+1) = -20/-5 = 4. This tells you that one of the partial fractions is 4/(x + 6). Now cover up (x + 1) and substitute, into what's left: x + 6). Now cover up (x + 1) and substitute, into what's left: -1, to discover that the other partial fraction is 1/(x + 1). Note: you can split partial fractions into four groups of problems: ones with: denominators of the form '(ax+b)(cx+d)', denominators with a repeated factor (i.e. [ax+b][cx+d]²), denominators with one quadratic factor (i.e. [ax+b][cx²+d]); the fourth type is a 'top-heavy' algebraic fraction, which occurs when the highest power of the highest power of the numerator is ... larger than or equal to the largest power of the denominator.
If the final fraction has a denominator of the form (ax+b)(cx+d) (let the numerator be f[x]), the partial fractions are of the form A/ax+b + B/cx+d, we just need to find A and B. Next, [] the two fractions add--, giving A(cx+d)+B(ax+b)/(ax+b)(cx+d); as it has the same denominator as f(x)/(ax+b)(cx+d), put f(x) = A(cx+d)+B(ax+b). To find A, put x = -b/a, then rearrange; to find B, put x = -d/c (then rearrange). Note: substitution of letters for their values is allowed: example: if A=3, Ax² + B(x-1) equals 3x² + B(x-1).
When the final fraction has a denominator of the form (ax+b)(cx+d)², write the partial fractions as A/ax+b + B/(cx+d)² + C/cx+d. Next, [] the fractions add--, and equate the top lines. You can then find values of the letters, by changing the value of x; but if that doesn't work for a variable, pick an x^n (advisably the [] one multiplied by the variable you seek) largest--, then replace each term with the amount of [] it has x^n's (thereafter rearrange) example: knowing A=2 and B=1, x²+12x-7 = A(x-1)²+B(x+2)+C(x+2)(x-1) gives 1=A+C: 1=2+C: C=-1; alternatively, using x, 5x²+5x+2=Ax²+C(x-2) gives 5=C.
If the denominator of the final fraction is in the form (ax+b)(cx²+d) (let the numerator be f), write the partial fractions as A/(ax+b) + Bx+C/(cx²+d). Next, [] the fractions add--, and equate the top lines. Finally, find the values of [...] A, B and C.
You may get an improper algebraic fraction f(x)/g(x) where the highest indice on f(x) is greater than or equal to the highest on g(x). In such a case, [] to get a [] and a [] divide to get a quotient (q) and a remainder (r), hence express f(x)/g(x) = q + r/g(x). Note: you may be able to split r/g(x) into partial fractions.
In general if one expands (1+x)^n where |x|<1, then the terms of the expansion get smaller and smaller, and so (often) can be ignored beyond the first [] terms three--. If n is negative or fractional, simplify what you want to expand to just (1+x)^n, in which case the expansion, that is, (1+x)^n = 1 + nx + n(n-1)x²/2! + n(n-1)(n-2)x^3/3! + ... . If the expansion given is (1+f[x]), where f(x) is not x, in the equation, replace x with f(x); in the expansion (1+f[x])^n (f[x] may equal x), the expansion, using the given formula, the expansion is valid for -1<f(x)<1. If the expansion is given as (a+x)^n, where a is an integer greater than 1, write the expansion as []*[] a^n * (1+[x/a])^n.
The movement from A to B may be given the vector AB (arrow overhead), while the movement from B to A may be given by the vector BA (arrow overhead). AB (arrow overhead) = BA (arrow overhead) * [] -1 --. A vector maybe given using three numbers (2 in 2D) in a column, generically (x,y,z); -(x,y,z) = (-x,-y,-z). Note: a single letter may be used to label a vector; the letter is often shown in lowercase and in bold type; YOU should write 'a' with an [] in place of 'a' to show that you mean a vector underbar--. To add one vector (a [underbar]) to another (b [underbar]), position them so that one [] where the other [] starts--finishes--, a+b can hence be represented by the vector from the start of a to the end of b.
The numbers used in a column vector are the c[] of the vector component--, adding vectors corresponds to adding their components.
The magnitude of a vector v, written as []v[] or []v[] |v| or ||v|| (so as to not confuse it with modulus), is the [] of the vector length--.
If the vector v extends into the region of the graph with negative values of x, and the angle between v and the negative x-axis is θ, then the angle between v and the direction of the positive x-axis is 180-θ. [#A40#].
Any lowercase letter may be used to label a vector, however, the use of i and j is reserved for two special vectors. i is a [] vector unit-- (i.e. it has length of 1) in the direction of the positive x axis. j is a [] vector unit--, in the direction of the positive y axis. In general, if v = (a,b), then (using i and j) v = ai + bj. Note: vector expressions in terms of i and j may be simplified in the usual way. In general, where k is a constant and v is a vector, kv is parallel to v, but k has to be [], because for vectors to be parallel, they have to have the same [] positive--direction--.
Let the origin be O: the vector from O to A (a) is the [] vector of A position--. Let there be another vector, but from O to B (b). Hence AB (arrow overhead) = b-a.
The unit vector in the direction of (vector) v is v/|v| (i.e. divide v by its magnitude). Example: v=3i+4j, |v|=5, thus the unit vector is v/5 = (3i-4j)/5. Note if you are asked to find a vector of magnitud a in the direction of v, multiply the unit vector (in the direction of v) by a.
The unit vector, the vector component in the third dimension is 'k'. If a = (x,y,z): (using the unit vectors) a = xi+yj+zk.
A vector that isn't fixed to the origin is called a free vector; a free vector has the same [] and [] as it would if it were a position vector magnitude and direction--, but its tail/start (which would have been the origin) can be anywhere.
If vector a = (x,y,z), |a| = √(x²+y²+z²).
In general if a and b are the position vectors of two points A and B, then the position vector of the midpoint of AB (M) is m = (a+b)/2 (=[b-a]/2 + a).
Let a = (xa,ya,za) and b=(xb,yb,zb) (a,b ss), then the angle θ, at the origin - between a and b, is given by cos θ = (xaxb + yayb + zazb)/|a|*|b|.
The scalar product of two vectors, a and b, is written as a.b; where θ is the angle between them, the scalar product is defined as the value of either |a||b|cosθ or xaxb+yayb+zazb (a,b ss); note: the former found through rearrangement of cos θ = (xaxb + yayb + zazb)/|a|*|b|, a = (xa,ya,za) and b=(xb,yb,zb) (a,b ss).
The scalar product of vectors a and b can be given by a.b = |a||b| cosθ. If θ=90°, then the vectors are perpendicular, at which point cosθ = 0. Hence a and b are perpendicular vectors if either [...] or [...] equal 0. |a||b|cosθ or a.b(=xaxb+yayb+zazb [a,b ss]).
In general, you can write any straight line in vector form if you know a [] on the line point--(p1,p2)/(p1,p2,p3), and a [] which is [] to the line vector--parallel--(d1,d2)/(d1,d2,d3). Then the equation is given in 2d by r = [...] and in 3d by [...] (p1,p2) + λ(d1,d2)--(p1,p2,p3) + λ(d1,d2,d3); changing [] will take you to any position, [], on the line λ--r--. Note: r is the [] [] of any point on the line position vector--.
(1,2,3 ss) If you have a line with equation of the form (p1,p2,p3) + λ(d1,d2,d3), and a point (a1,a2,a3), you can the value of λ at that point by equating a line (e.g. you can equate the first line to get p1 + λd1 = a1 and knowing p1, d1 and a1, deduce λ).
If a line passes through the x axis, then there must be a coordinate on the line of the form (a,0,0); with the y axis, there would be a coordinate on the line of the form (0,a,0); and with the z axis, there would be a coordinate of the form (0,0,a). To check that a line passes (for example) the x-axis, let []=[] (p1,p2,p3) + λ(d1,d2,d3) = (a,0,0), first equate the [] []s y coordinates, then rearrange for λ, then equate the [] [] and rearrange for [] z coordinates--λ--; if both values are the same, then there is a single value for which 'a' exists, thus it does pass the x-axis. Note: the technique is easily adaptable.
(1,2,3 ss) If the general vector equation of a line, r=(p1,p2,p3) + λ(d1,d2,d3), is given in terms of i, j and k, then the equation becomes r = p1i + p2j + p3k + λ(d1i+d2j+d3k).
With the vector equation of a line, the vector in the equation that's parallel to the line can go in either direction, the scalar (usually written as λ) can be changed to get to any point on the line. Thus (a,b,c) + λ(d1,d2,d3) and (d,e,f) + μ(nd1,nd2,nd3), where n is a constant, are parallel.
2D lines either [], are [] or are the [] line intersect--parallel--same--. However, in 3D, two lines can not parallel but will never intersect; they are called skew lines.
If you want to see if two 3D lines intersect: equate the [] and [] coordinates x and y coordinates, to get simultaeneous equations, then solve to find out the value of the scalars. Then equate the z coordinates, then [] in the [...] substitue in the value of the scalars. If the two sides are equal, then the lines intersect, otherwise, they don't. Note: if they intersect you can input the value of either of the scalars into the equation of the line it belongs to to get the coordinates of the intersection.
To find the angle between two lines: find the angle between the two directions, thus find the [] [] direction vectors.
Let there be a line on a graph - given a point, p, the shortest distance to the line is the length of the line [] which [] P to the line line segment--joins-- , that is, where the line segment is at [] [] to the line right angles--.
(1,2,3 ss) To find the shortest distance from the origin to a line r = (p1,p2,p3) + λ(d1,d2,d3). The first thing to do is find the point A, where the perpendicular line hits the line r. A point A is on the line r, its coordinates will satisfy the line's [] for some value of λ equation--λ--. Therefore the position vector of the point A must be of the form a = (p1+λd1,p2+λd2,p3+λd3). As a is perpendicular to the line r, the scalar product of their directions must be 0, hence [].[] = 0 (p1+λd1,p2+λd2,p3+λd3).(p1,p2,p3) = 0. Find the value of λ. To find the coordinates of A, substitute the value of [] into [] λ into r. The distance is the [] of the vector a magnitude--.
When trying to find the shortest distance from any point A(a1,a2,a3) to a line r=(p1,p2,p3) + λ(d1,d2,d3) (i.e. to a point B[b1,b2,b3]).You want to find the vector [] AB (arrow overhead). As B lies on line r, b = OB (arrow overhead) = ([]) (p1+λd1,p2+λd2,p3+λd3), for some value of λ. Thus AB = (p1-a1+λd1,p2-a2+λd2,p3-a3+λd3). Use the fact that AB .[] = 0 (d1,d2,d3)--, to find the value of λ, which is then to be inputed into AB. To get the distance, find AB's magnitude.
To convert the equation of a 2D line from vector to cartesian form, you can: add the [] and [] values seperately x and y--, and set them equal, respectively, to x and y to get two [] [] simultaeneous equation, then create an equation including [] and [] as the only variables (this can be achieved by getting rid of []) x--y--λ--.
To convert the equation of a line from vector to cartesian form: add the [], [] and [] values seperately x, y and z--, and set them equal, respectively, to x, y and z. Then, in each equation, rearrange to make λ the subject. Hence [] the three equations equate--, ignoring λ (such is the 3D cartesian equation of the line). Looking at the format, you can quickly write down the direction of the line: its components are the denominators (note: the denominator of a function of x is the x component etc.), and you can get a point on the line by, for each part, subtracting the variable and multiplying the rest by -1 (note: the part with the x variable maps onto the x-axis). If there are 0λ's for a certain line, write the others as usual then write it later, following an '[]' 'and' (e.g. x=2+3λ, y=7, z=4+5λ gives x-2/3=z-4/5 and y=7).
(1,2,3 ss) In general, a line in vector form written as r = (p1,p2,p3) + λ(d1,d2,d3) has cartesian form x-p1/d1 =y-p2/d2 = z-p3/d3.
For the vector equation of a plane you need the [] [] of a point on the plane position vector--(i.e.[a1,a2,a3]), and 2 non-[] [] [] which [] in the plane parallel direction vectors--lie-- (i.e. [p1,p2,p3] and [q1,q2,q3]). Where (a1,a2,a3) is the position vector of a point on the plane and (p1,p2,p3) and (q1,q2,q3) are non-parallel direction vectors lying in the plane: the vector equation of a plane is r = (a1,a2,a3) + λ(p1,p2,p3) + μ(q1,q2,q3).
To find where the line l meets the plane p (with the equations in vector form: equate the [], [] and [] values x, y and z--. Solve the consequent simultaeneous equations to find the value of the scalars. Having found the value of the scalar used in l, substitute it into l, to get the [] of the intersection points coordinates--.
To convert the equation of a plane to cartesian form: individually [] the values of x, y and z add--, and set them equal to, respectively, [...] x, y and z. To get three [] equations simultaeneous--. Then, eliminate the scalars (μ and λ). Then you can rearrange to get an equation of the form ax+by+cz+d=0.
(1,2,3 ss) Where a is the position vector of a point on the plane and n is a vector perpendicular to the plane, where n = (n1,n2,n3): the equation of a plane in cartesian form is given as n1x+n2y+n3z+d=0, where d = -a.n. Though sometimes, by using r as a general point (x,y,z) and knowing n, it is shortened to r.n=d.
When trying to find the intersection of a line with a plane (with its equation given in cartesian form): in the line, [] the values of x, y and z add--, and set them equal to, respectively, x, y and z. Then substitute those values, of x, y and z into the plane equation to find the value of λ. Finally put the found value of λ into the equation of the line, to get the point of intersection.
A plane has equation ax+by+cz+d=0. To find the shortest distance from the point A(x1,y1,z1) to this plane, we want to find the distance from A to the point where the [] line from A intersects the plane perpendicular-- (point P, which is also perpendicular to the plane, thus the distance AP). To do this we need to find P. Thus find the [] of the [] through A and P equation--line--, then see where it intersects the [] to get [] plane--P--. Now AP is perpendicular to the plane ax+by+cz+d=0; another vector perpenducular is [] = [] n = (a,b,c) (i.e. n1,n2,n3). Thus the line through A and P passes through []([]) and has direction [], hence the equation of the line is r = [] A(x1,y1,z1)--(a,b,c)--(x1,y1,z1) + λ(n1,n2,n3)--. Thereafter find where the line intersects with the plane. Now find the vector [] and hence [] AP (arrow overhead)--|AP*| *(arrow overhead)--.
To check if a point lies on a line (with equation in vector form): [] the line's equation and the position vector of the point equate--. Then find [] using the [] component λ--x--. Substitute the value of λ into the [] of [] and [] components of y and z. If the value works, then the point lies on the line.
To check if a point A(x1,y1,z1) lies on a line (with equation in cartesian form): substitute [] into the [] [] A--line equation--, taking x=[], y=[], z=[] x1--y1--z1--. If the equation is correct: the point lies on the line.
To check if a point A(x1,y1,z1) lies on a plane (with equation in cartesian form): substitute [] into the [] [] A--plane equation--, taking x=[], y=[], z=[] x1--y1--z1--. If the equation is correct: the point lies on the plane. (If you are given a plane with an equation in vector form, you can convert it to cartesian form).
If you want to see if a point lies on a plane: equate the [] and [] coordinates x and y coordinates, to get simultaeneous equations, then solve to find out the value of the scalars. Then equate the z coordinates, then [] in the [...] substitute in the value of the scalars. If the two sides are equal, then the point lies on the plane.
To see if a line lies completely within a plane, first show that they are parallel: if they are the direction perpendicular to the plane (n1,n2,n3) and the direction of the line (d1,d2,d3) will be perpendicular, thus have a [] [] of 0 scalar product--. Thereafter, show also that the line and the plane have a common point (i.e. show a point in the line lies on the plane/a point on the plane lies on the line).
If a line does not lie completely within a plane: then they are either just parallel or they [] at one [] intersect--point--.
Two planes are parallel if their perpendicular vectors are parallel. Two planes are perpendicular if their perpendicular vectors are perpendicuar.
The angle between two planes is the same as the angle between their perpendicular vectors.
Perpendicular vectors are also known as n[] normals.
|Parametric function| Parametric curves: in these cases, it is sometimes easier to write the equations for x and y in terms of another variable, t. To sketch the graph of x=f(t), y=g(t), for each value of t (usually within a given interval) work out the value of [] and [] x and y; these values pair up to be coordinates, which are then plotted. If f(t)=1/t and x or y = gf(t), then you may get a/multiple asymptote(s). Generally, due to the nature of the functions of t, the graph may only be valid for (a) certain interval(s).
To translate a parametric function by a vector (a,b) x=f(t) becomes [] and y=g(t) becomes [] x=f(t)+a--y=g(t)+b--.
To stretch a parametric function by a factor of a in the x direction, x=f(t) becomes x=af(t); to stretch a parametric function by a factor of a in the y direction y=g(t) becomes y=ag(t).
To reflect a parametric function in the y axis, x=f(t),y=g(t) becomes x=-f(t),y=g(t) ; to reflect a parametric function in the x axis, x=f(t),y=g(t) becomes x=f(t),y=-g(t).
To find where a parametric function crosses the x axis, make y= 0 and solve for t. The value(s) of t can them be put into the original equations.
To find where a parametric function crosses the y axis, make x= 0 and solve for t. The value(s) of t can then be put into the original equations.
To convert a parametric function to cartesian form, from x=f(t), get t= f^-1(x), then put that value of t into g(t), to get y=h(x). Alternatively, if y=g(t) is less complex, from there, get t= g^-1(y), then put that value of t into f(t), to get x=i(y), which can be rearranged to make y the subject. If given two parametric functions, x=f(t) and y=g(t), and asked to convert them to cartesian form: you are allowed to [], [], [] and [] them add, subtract, multiply and divide-- (i.e. treat them like [] equations simultaeneous--). Example: x = t + 1/t, y = t - 1/t: adding gives (1)x + y = 2t, subracting: x - y = 2/t; (1)*(2) = (x+y)(x-y)=4;x²-y²=4
The pair of parametric equations (using θ as the variable) that define a circle of radius 1, with centre at the origin is x=cosθ, y=sinθ. A circle with radius r, and centre (a,b) is given in parametric form as x=rcosθ+a, y=rsinθ+b, while in cartesian form it is given as (x-a)²+(y-b)²=r². An ellipse centre (0,0), with a x-radius of a and a y-radius of b, is given in parametric form as x=acosθ, y=bsinθ, while in cartesian form it's given as x²/a² + y²/b² = 1. Note: when trying to find the value of θ where the graph cuts an axis, be aware that there may be multiple solutions (e.g. sin θ = 0.5: θ=30°, θ=180°-30°).
With a parametric equation x=f(t), y=g(t), dy/dx = [], though it's usually written as dy/dt * dt/dx--dy/dt ÷ dx/dt--where dx/dx≠0--, this is through use of the chain rule.
To find dy/dx when x is a fuction of y (x=f[y]) or there are several occurrence of both x and y (or if one can't rearrange the equation to the form y=f(x) etc.), in these cases it may be best to differentiate implicitly. To do this, differentiate with respect with to x; afterwards you can rearrange for dy/dx; note: due to the chain rule - d(f[y])/dx = d(f[y])/dy * dy/dx.
When wanting to find the tangent or normal to a curve with an equation where y is not explicitly a function of x: in these cases typically: differentiate implicitly and find dy/dx as a function of x and y, and then get the values of x and y from a [] ([]) point (x,y), to find, at that point, the gradient. You may find the equation of the LINE with the equation: y-y1 = m(x-x1) (1 ss).
The trapezium rule: for a region with n strips of width h, we have: Area ≈ h/2(y0 + yn + 2(y1 + y2 + ... + y[n-1])).
|Integration by Inspection| If you want to integrate a*u^n, where u is a function of x, differentiate u^(n+1) (let the answer be b*u^n); you can hence deduce that the integral of a*u^n equals au^(n+1)/b. This method only works if the part multiplied by u^n (a) is a multiple of the [] of u^n differential--(i.e. the differential multiplied by a constant).
∫a/x dx = aln|x| +c; ∫e^(ax) dx = e^(ax)/a +c; ∫1/ax+b dx = ln|ax+b|/a +c; ∫f'(x)/f(x) = ln|f(x)| +c, hence, where h(x)=nf'(x), where n is greater or less than 1, ∫h(x)/f(x)= nln|f(x)| +c.
∫tan(x) dx = ln|sec x| +c
To make the integration of harder fractions easier: convert them to [] fractions partial--.
∫cos(ax) dx = sin(ax)/a +c. ∫sin(ax) dx = -cos(ax)/a +c. ∫sec²x dx = tan(x) +c; ∫cosec²x dx = -cot(x) +c.
d(cosec x)/dx = -cosec(x)cot(x); d(sec x) = sec(x)tan(x).
Integration of trigonometric functions (especially ones involving exponents greater than one) may be simplified using i[] identities.
With identities, if f(x)=g(x), f(ax)= g(ax).
If asked to integrate trigonometric functions of the form ∫sin(ax)cos(bx) dx: through use of addition formulae*, we can conclude that 2sinAcosB= sin(A+B)+sin(A-B). *sin(A+B)=sinAcosB+cosAsinB; sin(A-B)=sinAcosB-cosAsinB; sin(A+B)+sin(A-B) = sinAcosB+sinAcosB=2sinAcosB
Where 'a' is a constant, an integral of the form ∫1/(a²+x²) dx can be solved using the substitution x = atan(u); in general ∫1/a²+x² dx = tan^-1(x/a)/a +c.
Where 'a' is a constant, an integral of the form ∫1/√(a²-x²) dx can be solved using the substitution x = asin(u); in general ∫1/√(a²-x²) dx = sin^-1(x/a) +c. [#A41#]
∫f(x) dx = ∫(f[x]* dx). Example: ∫cos²x dx/cosx = ∫cosx dx.
When integrating some products, inegration by parts may need to be done more than once. If you have integrated a product to get the integral, which includes one term that is the integral of a product, on that term, you can use integration by parts. With integration by parts: ∫uv' dx = uv - ∫vu' dx.
If I is the integral you want, and I = f(x) + aI, simply rearrange to find the value of I.
An equation which involves a derivative (e.g. dy/dx) is called a differential equation.
If asked to solve an equation of the form dy/dx = f(y), to start, put all the []-terms on [] side y--one--. Then [] both sides integrate-- with respect to x (considering that dy/dx * dx = dy). Afterwards, make the subject y. Note: where c is the arbitrary constant: f(c) can be replaced with another representation of a constant (e.g. A). This method is known as the method of seperation of variables. Example: dy/dx = e^-2y;... ∫e^2y dy = ∫ dx: e^2y/2 = x +c; e^2y = 2x + A; y = ln(2x+A)/2. Note: c can be replace by f(c), where f(c) can yield any constant in the appropriate range (e.g. in some cases c may be replaced by ln c).
To solve a differential equation, get rid of the derivative.
For any growth or decay problem, the solution to dy/dx = ay, where 'a' and 'k' are constants, is y = ke^(ax); if a<0, then it's a [] problem decay--, and if a>0, then it's a [] problem growth--.
(1,2 ss) If asked to find the particular solution for a differential equation for which y=y1 and x = x1; after getting rid of the derivative, to get an equation with y and x as terms: input y=y1 and x=x1 to find (usually) c (the arbitrary constant) (,but it could be any other algebraic term in the equation).
When integrating, with integrals, it is permitable to have the arbitrary constant on one side (e.g. ∫f(x) dx = ∫g(x) dx +c).
If dN/dt is proportional to N, then dN/dt = kn, where k is a constant.
With differential equation (esp. real life examples): let the two variable be x and y, to start, it is usually advisable to solve the differential equation; rearranging usually puts it in the form y=Ae^kx, where A and k are constants. A (though it can be done for k) is often found by substituting in, at a certain point, the values of x and y (usually where x=0). The other constant, which has not been found, can be found through another substituton (of the values of x and y).
Created by: Toluo

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