Solve: у" + 5у'" — 2y' - 24у — 42е 5t 2y' — 24у 3D 42e — 2, у'(0) — — 6, у'"(0) —D — 30 | | | y(t) =

Q: c) Find i. y' for x4 + y = 17 If x = 3 and y = 3, calculate y'

A: Here is the solution of above problem.

Q: find extreme Value of function fex, y) = x°t 3xy?-3x²-3y=t7 1

A: Given:  fx,y=x3+3xy2-3x2-3y2+7 ⇒fx=3x2+3y2-6x, fy=6xy-6y To find extreme points put fy=0, fx=0…

Q: 4X-) dx dy y2 ret -1

A: Given: ∫12∫-10ex2y-4x-1y2dxdy

Q: Find if x'y+y' = x dy dx A) 2x+2y 1-2xy-2y B) x2 1-2ху C) x2+2y D) O A O B O c O D

A:

Q: -t,x-1 dt

A: Please see the explanation below

Q: дz If z = f (x, y), find and for the following equation ду дz дх

A:

Q: Find mZX (3y-71° X Y (2x) O W (x+181° (3y + 1)º

A: Solve for the angle X

Q: Q1- Find the maximum vate of change of Rexy) t -ツ fcxy) = Xe fCxg) = Xe at the point (e,o) %3D

A: Given problem:- Find the maximum rate of Change f(x, y) = xe-y at point (2,0)

Q: 4. (D2 – 5D + 6)y = 15sinx – 15cosx -

A:

Q: Find Duf at P. f(x, y, z) = 6x³y³z³; P(2, — 2, 2); u = Duf = i + - 2|3 k

A:

Q: 2-у (бх + 2у?)dx dy -2

A: On solving this we will get

Q: + y = 2. What is y" at the point (1, 1)? y ² Suppose that x +

A:

Q: Find the distance between x = 10 -3 and y = ㅗ -5

A:

Q: Eliminate the arbitrary constant. y=Ax³+Bx³ (A B 7xy" - x²y + 15y = 0 D x²y '- 7xy" + 15y = 0 Cx²y"…

A: To find- Eliminate the arbitrary constant. y = Ax5 + Bx3

Q: Lt² V4-t² dt V(4-x)/(2\x) x(4-x)

A:

Q: 2. fote-t² dt

A:

Q: Find y": Vy - Vx = Va

A:

Q: If fxy (x, y) = , ,0 2) =

A: Given,fXY(x,y)=x+y27 ; 0<x<2 , 3<y<5

Q: The roots of the characteristic equation of y"-7y=0 are:  a. √7.-√7  b. 0, 7 c. 7,7 d. 7.-7

A: Here I am attaching image so that you understand each and every step.

Q: 3) Find y"(0), where y = 4x + 2x + 3

A: A function is a relation which maps each value in its domain to exactly one value in its range. A…

Q: A 4,-1) B(4,-2) find lime x approachesto1 from postve dinection.

A:

Q: Q: (2x+y)(x+y) ах Find -?

A: Given query is to find indicated partial derivative.

Q: particle started at A(1,0) circled the origin once  an d returned toA(1,0) .what were the change in…

A: Given: A particle started at A(1,0) circled the origin once  and  returns to A(1,0)

Q: If B = x°ya, + (2x + y)a, – (y – 2)a, find (a) V B (b) V X B (c) V (V B) (d) V x V X B

A: Given : B=x2yax +(2x2+y)ay - (y-z)az

Q: 2. Sx³(a² + x²)° dx ? + x?)% dx

A: Using property of integration,