Y=Ae^x+Bxe^x Solve differential equation?
1. Y=Ae^x+Bxe^x Solve differential equation?
Step-by-step explanation:
There are many differential equations where y=Ax2+Bxex is a solution, although some of them might have other solutions as well.
On our first attempt, we can try to find a linear homogeneous equation with constant coefficients. Usually, when we are given such a problem, we set up a characteristic polynomial equation, then solve it, and finally we use the roots to this equation to find solutions to the differential equation.
But here, we’ll work backwards by finding roots of characteristic equation first, and then finding the characteristic equation, and finally the differential equation.
If characteristic equation has real root r with multiplicity n, then general solution is:
y=erx(c1+c2x+⋯cnxn−1)
So we need to find r and smallest value of n that will fit each of the two linearly independent solutions.
y=Ax2=Ae0xx3–1⟹r=0,n=3⟹r3=0
y=Bxex=Be1xx2–1⟹r=1,n=2⟹(r−1)2=0
Therefore characteristic equation has roots r3 and (r−1)2
r3(r−1)2=0
r3(r2−2r+1)=0
r5−2r4+r3=0
The differential equation that matches characteristic equation is:
y(5)−2y(4)+y(3)=0
Unfortunately, this differential equation has general solution:
y=c1x2+c2x+c3+c4xex+c5ex
If you need to eliminate the unnecessary terms, you can use initial conditions, such as:
y(0)=0
y(3)(0)=3y′(0)
y(4)=43y(3)
Plugging into WolframAlpha (with initial conditions), we get:
Wolfram|Alpha
y=13x(c1ex+c2x)=c23x2+c13xex=Ax2+Bxex
However, if we want to find a differential equation that only has the two linearly independent solutions shown, and without using initial conditions, we’ll need to find a second order differential equation. And each time we differentiate, we’ll need to eliminate one of the constants.
y=Ax2+Bxex⟹Ax2=y−Bxex
Differentiate:
y′=2Ax+Bxex+Bex
xy′=2Ax2+Bx2ex+Bxex
Eliminate constant A:
xy′=2(y−Bxex)+Bx2ex+Bxex
xy′=2y+Bx2ex−Bxex⟹Bex=xy′−2yx(x−1)
Differentiate again:
xy′′+y′=2y′+Bx2ex+2Bxex−Bxex−Bex
xy′′−y′=Bex(x2+x−1)
Eliminate constant B:
xy′′−y′=xy′−2yx(x−1)(x2+x−1)
x2(x−1)y′′−x(x−1)y′=x(x2+x−1)y′−2(x2+x−1)y
x2(x−1)y′′−x(x2+2x−2)y′+2(x2+x−1)y=0
y=c21−x−−−−−√x2x−1−−−−−√+c1ex1−x−−−−−√xx−1−−−−−√=c2ix2+c1ixex=Ax2+Bxex
2. eliminate the arbitrary constant on y=ae^x +bxe^x
Final Answer:
y" - 2y' + y = 0
3. IN 2x - 4 * +3 2x - 4 = (1 pt) X= (3pts) AE = AN + EN AE = + 2x - 4 (1 pt) AE = (1 pt) PA = x + 3 PA= (1 pt) PA = PE PE = (1pt) CRITERIA Correct answer with accurate complete solution. Correct answer. No attempt. Page 2 of 3
Answer:
EN= 2x-4
AN=6
PE=X+3
.EN=AN
2x-4=2(x-4)(x-4)
=2x-8+2x+8
=2x+2x-8+8
4x=-8
x=2
AE=AN+EN
AE=6+2x-4
AE=2x-4-6
PA= x+3
PA= x=-3
PA=PE
PE=x+3
good luck!!Sana nakatulong
4. The figure below is a parallelogram. The diagonals AC and BD intersect at E. If AE = 2x and EC =12, what is the value of x
Answer: x is equal to 6
Step-by-step explanation:
Use the property "diagonals bisext each other"
5. A. Identify if the following is an Algebraic Expression (AE), an Equation (E), or an Inequality (IE). Write AE if it is an algebraic expression, E if an equation and IE if an inequality. Write it on the space provided before the number.1) 2x + 1 =52)3y + 6 + 5x3)3x + 2 ≥ 84)5y = 155)4x < 126) 3 ( x^2+2x-1)7) X ≤ 28) x + 1 > 109)5x – 1 = 2x + 310) 1/3 x- 2/5 y+1
Answer:
1) 2x + 1 =5
E
2)3y + 6 + 5x
AE
3)3x + 2 ≥ 8
IE
4)5y = 15
E
5)4x < 12
IE
6) 3 ( x^2+2x-1)
AE
7) X ≤ 2
IE
8) x + 1 > 10
IE
9)5x – 1 = 2x + 3
E
10) 1/3 x- 2/5 y+1
AE
Remember:
- AE: only numbers, variables, and operations
- E: has equal symbol
- IE: has >, < symbols
6. If AE = 2x + 10 and CE = 6x – 30, find AC.
Answer: AC=60
Step-by-step explanation: If AE is equal to CE (The midpoint is E)
So the equation is
2x+10=6x-30
2x-6x=-10-30
-4x=-40
x=10
then
2x+10
2(10)+10
20+10
=30
then
AC is the diagonal so 30x2 is
60
AE=30
CE=30
Sorry in advance if this has mistakes/wrong answers but this is my opinion/answer Your welcome in advance
Answer:
AE=2×+10*AC*by*get*step^2*tha
Step-by-step explanation:
is that the anwer
7. ABCD is a parallelogram. if AD =(2x - 10) cm and BC=(x + 30) cm, then what is the length of BC?Base angles of an isosceles trapezoid areABCD is a parallelogram. The diagonals AC and BD intersect at E. if AE =2x and EC =12, what is x?LIKE is a kite with LI =~ IK, and LE =~ KE. if LI =(2x) cm and IK =(x + 5) cm, how long is LI?Maayos na sagot po ang kailangan. Salamat!
Answer:
1. BC = 70
AD = BC(2x-10) = (x+30)*transpose*2x - x = 30 + 10x = 40*substitute*BC = (x+30)BC = 40 + 30BC = 702. x = 6
2x = 12(1/2)2x = 12(1/2)x = 63. LI = 10
2x = x + 5*transpose*2x-x = 5x = 5*substitute*LI = 2xLI = 2(5)LI = 108. ACTIVITY 2. ABCD is a parallelogram. Solve each D E A 8 1. If AE = 2x + 10 and CE = 6x - 30. Find AC. 2. If AD = 2(x - 3) and BC = 7(x -9). Find AD + BC. 3. If DE = 3x - 15 and BE = 2x + 10. Find BD. 4. If DC = 4x - 5 and AB = 2x + 15. Find AB + DC. == .
Answer:
the answer is in the picture
Step-by-step explanation:
pa brainliest nalang po thank you
9. 3.In ABC, DE || BC, | AD | = (2x), |AB| =20 cm, |AE| =(3x) and |EC| =12 cm. Find |AD|
Answer:
The answer is in the picture.
[tex]\color{skyblue}{Hope \: it \: helps}[/tex]
10. 5. If CS=2x – 3 and AS = 3x – 11, find FC. 6. If ES = 3x + 5 and FS = 2x - 3, find EA. - 7. If ES = 2x – 3 and CF = 3x +8, find ES and CF. = - 8. If CS = x +12 and AE = 4x – 12, find CS and AE. = 9. If AE = 41, what must be the length of chord EF so that the perimeter of AEFS is 50? 10. If CF = 50, what must be the length of chord AC so that the perimeter of AACS is 60?
Answer:
5. If CS=2x – 3 and AS = 3x – 11, find FC. 6. If ES = 3x + 5 and FS = 2x - 3, find EA. - 7. If ES = 2x – 3 and CF = 3x +8, find ES and CF. = - 8. If CS = x +12 and AE = 4x – 12, find CS and AE. = 9. If AE = 41, what must be the length of chord EF so that the perimeter of AEFS is 50? 10. If CF = 50, what must be the length of chord AC so that the perimeter of AACS is 60?
11. 17. Given the O B with BD I AE. AC = 4x – 40 and CE = 2x + 12. Find: 2. AE
. 4. ^2 ∙ ^5=^2+5 *
a. Product Law
b. Power of Power Law
c. Power of a Product Law
d. Quotient Law
5. (^2)^5=^2∙5 *...
https://brainly.ph/question/22960092?utm_source=android&utm_medium=share&utm_campaign=question
12. 2. Given:BD | AE. Find the ratios of the following and solve for the value of x . x of 4 10 2x 8,
Answer:
folding it will give the answer
Step-by-step explanation:
69
13. AE=2x + 4 while CE= x + 20 how long us CE?A. 24 cmB. 30 cmC. 36 cmD. 42 cm
Answer:
c.36 cm
Explanation:
Sana makatulong
14. 24. PLAY is a parallelogram. The diagonals PA and LY intersect at E, if AE = 2x and EP = 12, what is x and PA?
Answer:
take note:
AE = EP
AE + EP = PA
finding x.
AE=EP
2x=12
x=6
finding PA
since AE and EP are equal, then simply multiply EP by 2 so,
PA=24
Step-by-step explanation:
hope it helps
15. For numbers 14 and 15, consider parallelogram ABCD Given: AE = (2x + 10) cm CE = (6x - 30) cm A B 14. Find AE? a. 30 b. 40 c. 50 d. 60 15.Find AC? c. 50 d. 60 a. 30 b. 40
Answer:
14. B. 40
15. A. 30
@CallMeMentari~
Answer:
14.B
15.A
Step-by-step explanation:
Sana makatulong mga idok
16. Given the O B with BD I AE. AC = 4x -40 and CE = 2x + 12. Find; B. 4. AG 5. AE A dC E
Answer:
4. 64
5. 64
Step-by-step explanation:
4x-40=2x+12
4x-2x=12+40
2x/2 = 52/2
x=26
4. 4(26)-40
=64
5. 2(26)+12
=64
17. 1. Eliminate the arbitrary constants on the given equation: y/e^2x=C1cos3x+C2sin3x 2. Eliminate the arbitrary constant on the given equation: r=Ae^−6θ+Bθe^−6θ
Step-by-step explanation:
which term of the following arithmetic secqence
18. If AE = 3x+10 and CE = 2x+30, find AC.
Answer:
if a quadrilateral is a parallelogram then what is true to it's consecutive angles
19. 1.If AE=y,then AC=___2. If m/_ABD=43⁰ ,then m/_CBD =___3. If AD =8 and BC = 2x+ 4,then x =_____ 4. If m/_ ABC = 110⁰ and m/_BDC =56⁰,then m/_ DBC=_____ 5.If DE= 2x + 2 and EB = 5x - 10 ,then x=_____
Answer:
1.if AE=y,then AC=X
2.
3.
4.
5.if DE=2x+2 and EB=5x-10, then X= 2²
20. 7. ABCD is a parallelogram. The diagonals AC and BD intersect at E. If AE = 2x andEC = 12, what is x?A. 5B. 6C. 7D. 8
Answer:
D.8
Step-by-step explanation:
it so ez napakadali nagtatanong Ka pa
21. en: If quadrilterals SAVE is a rectangle, and SV = 5x - 8 and EO = 2x + 3.SAEV5. Find the value of x.6. What is the length of AE?
The right bowling ball would have greater kinetic energy than the left bowling ball. This is because kinetic energy is directly proportional to the square of the ball's speed.
\texttt{ }
Further explanation
Given:
m₁ = m₂ = m
v₁ = 3 m/s
v₂ = 5 m/s
Asked:
Ek₁ : Ek₂ = ?
Solution:
Ek_1 : Ek_2 = \frac{1}{2}m_1 (v_1)^2 : \frac{1}{2}m_2 (v_2)^2Ek
1
:Ek
2
=
2
1
m
1
(v
1
)
2
:
2
1
m
2
(v
2
)
2
Ek_1 : Ek_2 = \frac{1}{2}m (3)^2 : \frac{1}{2}m (5)^2Ek
1
:Ek
2
=
2
1
m(3)
2
:
2
1
m(5)
2
Ek_1 : Ek_2 = 3^2 : 5^2Ek
1
:Ek
2
=3
2
:5
2
Ek_1 : Ek_2 = 9 : 25Ek
1
:Ek
2
=9:25
From the results above, we can conclude that the right bowling ball would have greater kinetic energy than the left bowling ball. This is because kinetic energy is directly proportional to the square of the ball's speed.
\texttt{ }
Learn more
Force and Motion : https://brainly.ph/question/2155739
Acceleration : https://brainly.ph/question/589534
Total Work : https://brainly.ph/question/2501947
\texttt{ }
#LearnWithBrainly
22. if AE=(5x-5) cm, and CE= (2x+7)cm, find x and AC
Step-by-step explanation:
AE = 5x - 5
CE = 2x + 7
Subtract AE and CE
0 = 7x -12
Add 12 to both sides
12 = 7x
Divide both sides by 7
x = 12/7
AC = AE - CE
= (5x - 5) - (2x + 7)
= 3x - 12
= (3 x (12/7)) - 12
= 24/7 - 12
= 12/7 cm
23. Using Rectangle CARE, CR = 2x and AE = x + 5. What property/theorem of rectangle will be used to identify the measure of the diagonals
Answer:
triangle
i dont shink tama
24. The figure is a parallelogram. The diagonals AC and BD intersect at E. if AE = 2x and EC = 12, what is x?
Answer:
In a parallelogram, the diagonals bisect each other.
In a parallelogram, the diagonals bisect each other.Hence,
In a parallelogram, the diagonals bisect each other.Hence,AE = CE & BE = DE
In a parallelogram, the diagonals bisect each other.Hence,AE = CE & BE = DE2x = x + 2
In a parallelogram, the diagonals bisect each other.Hence,AE = CE & BE = DE2x = x + 2x = 2
In a parallelogram, the diagonals bisect each other.Hence,AE = CE & BE = DE2x = x + 2x = 2y + 10 = 4y - 8
In a parallelogram, the diagonals bisect each other.Hence,AE = CE & BE = DE2x = x + 2x = 2y + 10 = 4y - 83y = 18
In a parallelogram, the diagonals bisect each other.Hence,AE = CE & BE = DE2x = x + 2x = 2y + 10 = 4y - 83y = 18y = 6
In a parallelogram, the diagonals bisect each other.Hence,AE = CE & BE = DE2x = x + 2x = 2y + 10 = 4y - 83y = 18y = 6Hence x = 2 and y = 6
Step-by-step explanation:
ihope it helps
Answer:
The figure is a parallelogram. The diagonals AC and BD intersect at E. if AE = 2x and EC = 12, what is x?6
Step-by-step explanation:
please heart rate follow brainliest (^^)
25. if ae=x+7 be=x+1 ce =x+3 and de=2x - 1 find x.
Answer:
Applying basic proportionality theorem, we getADDC= BEEC 2xx+3 = 2x−1x (x+3)(2x−1)=x(2x)2x 2 −x+6x−3=2x 25x=3x= 53Step-by-step explanation:
>.<?26. given the circle B with BD T AE. And AC=4x-40 and CE=2x+12. Find AC
4x - 40 = 2x + 12
2x = 52
x = 26
AC = 4*26 - 40
AC = 64
make this brainliest if you want me to help you more :)
27. (pls i need serious answers don't answer if u dont know I'll report u fr)18. Quadrilateral CDEF is a parallelogram. Find x and y if CG=x+2y, EG-4, FG = 3x-y, and DG = 5. FE. Quadrilateral ABCD is a rhombus. Its diagonals AC and BD intērsect at A D Η | E FE. Quadrilateral EFGH is a rectangle. 19. If AE = 3x+3 and CE = 5x-3, find AC.20. If DE = 2x+2 and BE = 3x-1, find BD. G C F 21. If EF = 4x-12, HG = 2x + 6, and FG = 2x, find its perimeter.22. If HE = x + 6, Gㅋ = 3x - 2, and HG= 2x + 6 find it's area
Answer:1.x = y = 2.5 and FG = 5
2.AC = sqrt(34x^2 - 12x + 18)
3.BD = sqrt(13x^2 + 2x + 5)
4.EFGH is 12x - 18
5.EFGH is -4x^2 + 4
28. ae exponent is D Learning Task 1 Simplify by writing in exponential form 1. 2(2)(2)(2)(2)(2) 26 Write in expanded form (2x)4 = (2x) (2x) (2x) (2x) 2. (За)(За)(За)(За) = (xy)7 = 3. (-4)(-4)(-4)(-4)(-4) = (-ab)6 = Exponents have its own rule in performing mathematical operations.PA SAGOT PO PLSS UNG MAAYOS SABA
Answer:
11² 3⁴a-4⁵Step-by-step explanation:
im sure is correct
29. IF AE=2X+10 and CE=6x-30 find the AC
Answer:
AC=60
Step-by-step explanation:
If AE is equal to CE (The midpoint is E)
So the equation is
2x+10=6x-30
2x-6x=-10-30
-4x=-40
x=10
then
2x+10
2(10)+10
20+10
=30
then
AC is the diagonal so 30x2 is
60
AE=30
CE=30
Answer:
AC=60
Step-by-step explanation:
If AE is equal to CE (The midpoint is E)
So the equation is
2x+10=6x-30
2x-6x=-10-30
-4x=-40
x=10
then
2x+10
2(10)+10
20+10
=30
then
AC is the diagonal so 30x2 is
60
AE=30
CE=30
30. quadrilateral abcd is a parallelogram . the diagonals ac and bd inrersect at e. if ae = 2x and ec = 12 what is x?
Answer:
6
Step-by-step explanation:
Both AE and EC are on the same line.
AE=EC
2x=12
x=6
