-7х - 2у %3D —13 2y + 11 %3D

Q: y"+y' - 2y = x2

A: Solution :-

Q: y(6) –4y(5) +3y/) =-27 %3D

A: The given differential equation is y6-4y5+3y3=-27. We have to find the solution to the given…

Q: 5p2 900 = 0 %3D

A:

Q: у" -2 у' + у 3D0 0

A:

Q: 7x-2 b) xp: 3x-1

A:

Q: y" -4y' +4y = te2t

A: Consider the following different differential equation: The solution is divided into two sets,…

Q: y' = y2 - 4

A:

Q: y" - 4y' = 0

A: Given: Ordinary equation is  y''-4y'=0

Q: 2) y" - y' — 6y = 0. бу

A: Follow the procedure given below.

Q: у" - у' — 2у %3D —4

A: To solve the given ODE.

Q: Simplify ^y2 - 10y + 25

A: Given expression is :                                             y2-10y+25     We can factorize…

Q: y" + 5y' - 4y = 0

A: SOLUTION-

Q: (c).xv-12-2≦y2

A:

Q: y" + 4.5y' = 0 %3D

A:

Q: y" - 4y' + 5y = 0

A:

Q: 1) y" - 36y = 0.

A:

Q: (c) 2y+1 - 5yk+1Yk + 2y% = 2,

A:

Q: 2y2+12y-54

A:

Q: -2 r. -2 |= -5 %3D and

A:

Q: 4x-7y=28

A:

Q: – 2ay + i(bx -

A:

Q: (b) y" +y + y³ = 0

A: Given DE is y''+y+y3=0 to find the equilibrium point, we have to convert it ti system of first order…

Q: y?V63x5y4

A:

Q: fcx)= V16-x?

A: Answer

Q: 4y - = 23+

A:

Q: y" -2y'+5y=0

A: Given differential equation, y''-2y'+5y=0

Q: y = -x – 6 - 2х 3D — Зх + 6 4)

A: NOTE: Refresh your page if you can't see any equations. . the first equation is  the second…

Q: 3X-1 xp +4

A: Follow the procedure given below.

Q: - y" - 5y + 4y= 2et

A: y'' -5y' +4y=2etThe general solution to axy''+bxy'+cxy=gx can be written asy=yh+ypyh is the solution…

Q: y' - 4y = 4y18

A:

Q: (y>-- (2x-y 2-3 AALAM

A: Here we have to graph the given inequality:-  y>-x2-22x-y ≥-3

Q: 3y-6(y+z)+y

A: Distribute -6.

Q: 4) y>-4x A) B) C) D)

A:

Q: | (27y - 8x) . R

A:

Q: y′ = 1 + 2y

A: Given,

Q: y = x² - 4

A: as per guideline i can solve only first question

Q: y = |x + 4|

A: Given: - The equation is, y=x+4 Tp sketch: - The graph of the given equation by point plotting

Q: y = x² + 1 y = 2r- 2

A: Here, from the given figure, the area bounded by the given curves are: A=∫-12∫2x-2x2+1dy dx

Q: y' – 2y = 3e'

A:

Q: 2x5y8 h) 8y5

A:

Q: (-3b)(-4)

A: To evaluate(-3b)(-4)

Q: b. y = -x² + 4 y 2 2

A: SOLUTION-

Q: y" + 3y' = 6 %3D

A:

Q: y′ = −1 − 2y

A: Plug y' = dy/dx then separate the variables and integrate both sides.

Q: (3x')* (Зх (x'y')²

A: (3x3)4x1y42=3×3×3×3×x12x2×y8

Q: - (D-2D'+5) 2 = 0

A:

Q: + 2t y' - 6y = 0

A:

Q: y" + 2y' - y = 0

A: Definition used -         The solution of the linear differential equation -         The…

Q: +y" - 4y +6y= 3-3

A:

Q: y = x² + x y = x + 5 2

A: y=x2+x y=x+5   a2-b2=(a-b)(a+b)