Answer:y=
Acosx+Bsinx +cosx ln(cosx)+x sinx
Step-by-step explanation:
given equation y''+y=secx
auxiliary equation
so CF is y=Acosx+Bsinx
now
using wronskian formula
=
now f(x)=secx
now particular integrals are
PI=cosx ln(cosx)+x sinx
total solution
y= C.F+P.I
y=Acosx+Bsinx +cosx ln(cosx)+x sinx
<span>Vector Equation
(Line)</span>(x,y) = (x,y) + t(a,b);tERParametric Formx = x + t(a), y = y + t(b); tERr = (-4,-2) + t((-3,5);tERFind the vector equation of the line passing through A(-4,-2) & parallel to m = (-3,5)<span>Point: (2,5)
Create a direction vector: AB = (-1 - 2, 4 - 5)
= (-3,-1) or (3,1)when -1 (or any scalar multiple) is divided out.
r = (2,5) + t(-3,-1);tER</span>Find the vector equation of the line passing through A(2,5) & B(-1,4)<span>x = 4 - 3t
y = -2 + 5t
;tER</span>Write the parametric equations of the line passing through the line passing through the point A(4,-2) & with a direction vector of m =(-3,5)<span>Create Vector Equation first:
AB = (2,8)
Point: (4,-3)
r = (4,-3) + (2,8); tER
x = 4 + 2t
y = -3 + 8t
;tER</span>Write the parametric equations of the line through A(4,-3) & B(6,5)<span>Make parametric equations:
x = 5 + 4t
y = -2 + 3t ; tER
For x sub in -3
-3 = 5 + 4t
(-8 - 5)/4 = t
-2 = t
For y sub in -8
-8 = -2 + 3t
(-8 + 2)/3 = t
-2 = t
Parameter 't' is consistent so pt(-3,-8) is on the line.</span>Given the equation r = (5,-2) + t(4,3);tER, is (-3,-8) on the line?<span>Make parametric equations:
x = 5 + 4t
y = -2 + 3t ; tER
For x sub in 1
-1 = 5 + 4t
(-1 - 5)/4 = t
-1 = t
For y sub in -7
-7 = -2 + 3t
(-7 + 2)/3 = t
-5/3 = t
Parameter 't' is inconsistent so pt(1,-7) is not on the line.</span>Given the equation r = (5,-2) + t(4,3);tER, is (1,-7) on the line?<span>Use parametric equations when generating points:
x = 5 + 4t
y = -2 + 3t ;tER
X-int:
sub in y = 0
0 = -2 + 3t
solve for t
2/3 = t (this is the parameter that will generate the x-int)
Sub t = 2/3 into x = 5 + 4t
x = 5 + 4(2/3)
x = 5 + (8/3)
x = 15/3 + (8/3)
x = 23/3
The x-int is (23/3, 0)</span>What is the x-int of the line r = (5,-2) + t(4,3); tER?Note: if they define the same line: 1) Are their direction vectors scalar multiples? 2) Check the point of one equation in the other equation (LS = RS if point is subbed in)What are the two requirements for 2 lines to define the same line?
Answer:
y = -x - 2
Step-by-step explanation:
Find two points on the line
(0, -2)
(4, -6)
(Y2-Y1)/(X2-X1) = gradient of the line
Where (X1,Y1) = (0, -2) and (X2,Y2) = (4, -6)
(-6--2)/(4-0)
(-6+2)/4
(-4)/(4)
-1
Y = (-1)X + c
Substitute either set of coordinates in to find c.
(0, -2)
Y = (-1)X + c
-2 = (-1)0 + c
-2 = 0 + c
-2 = c
(4,-6)
Y = (-1)X + c
-6 = (-1)4 + c
-6 = -4 +c
-6 + 4 = c
-2 = c
Y = -X -2
Answer:
22900
Step-by-step explanation:
multiply 350 by 6 and get 2100
subtract 2100 from 25k and get 22900
