Question

find y' for the function

1. (y-2)^7=3x^2+2x-2

2. 3y^3+2x^3=3

3.(4y^2+3)^4+3x^5-5=0

4. 4x^2+3x^2y^2-y^3=3x

Answer #1

Use
Gaussian Elimination to solve and show all steps:
1. (x+4y=6)
(1/2x+1/3y=1/2)
2. (x-2y+3z=7)
(-3x+y+2z=-5)
(2x+2y+z=3)

Solve for the general solution
x^4y''''+4x^3y'''+3x^2y''-xy'+y=0

solve the system of equations
1: y=3x^2-2x-1
2x+3y=2
2: x^2+(y-2)^2=4
x^2-2y=0

Solve the system using 3x3
3x-2y+z=2
5x+y-2z=1
4x-3y+3z=7

Consider the linearly independent set of vectors
B= (-1+2x+3x^2+4x^3+5x^4, 1-2x+3x^2+4x^3+5x^4,
1+2x-3x^2+4x^3+5x^4, 1+2x+3x^2-4x^3+5x^4, 1+2x+3x^3+4x^3-5x^4)
in P4(R), does B form a basis for P4(R) and why?

Minimize C = −2x + 3y subject to 3x + 4y ≤ 24, 7x − 4y ≤ 16, and
x, y ≥ 0.

Solve the system of equations using an inverse matrix
-4x-2y+z= 6
-x-y-2z= -3
2x+3y-z= -4
Choose one:
a. (-1, 0, -2)
b. (1, 0, -2)
c. (1, 0, 2)
d. (-1, 0, 2)

5. The planes: 2x − 3y + z = 1 and 3x − 2y − z = 0 …
(Explain/Show your Work)
a. Are parallel
b. Are Coincident
c. Meet in a Line
d. Meet at a Point

Solve the IVP using the Eigenvalue method.
x'=2x-3y+1
y'=x-2y+1
x(0)=0
y(0)=1
x'=2x-3y+1
y'=x-2y+1
x(0)=0
y(0)=1
Solve
the IVP using the Eigenvalue method.
x'=2x-3y+1
y'=x-2y+1
x(0)=0
y(0)=1

Verify that the given function is the solution of the initial
value problem.
1. A) x^3y'''-3x^2y''+6xy'-6y= -(24/x) y(-1)=0 y'(-1)=0
y''(-1)=0
y=-6x-8x^2-3x^3+(1/x)
C) xy'''-y''-xy'+y^2= x^2 y(1)=2 y'(1)=5 y''(1)=-1
y=-x^2-2+2e^(x-1-e^-(x-1))+4x

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