Question

x^3y’’’+3x^2y”+2xy’=0

Answer #1

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(2xy+x^2+3y^2)dy/dx+(y^2+2xy+3x^2)=0

x^3y''' - x^2y'' - 3xy' =3x

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

3. Consider the equation (3x^2y + y^2)dx + (x^3 + 2xy + 5)dy =
0. (a) Verify this is an exact equation
(b) Solve the equation

7) Solve x^2(x + 1)y’’ – 2xy’ + 2y = 0

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

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 differential equation
(2y^2+2y+4x^2)dx+(2xy+x)dy=0

solve for x(t) and y(t):
x'=-3x+2y;
y'=-3x+4y
x(0)=0,y(0)=2

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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