Question

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

Answer #1

7) (1 − x2)y'' − 2xy' + 2y = 0; y(0) = 0, y'(0) =
3
Find the power series solution about x=0.

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

1) Solve by power series around x=0: y"-2xy'-2y=0
(Find the first three nonzero terms of each of the LI
solutions)

solve ivp: (y+6x^2)dx + (xlnx-2xy)dy = 0, y(1)=2, x>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

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

consider ivp given by x^2y" + 2xy' - 6y = 0 w/ y(1) = 1, y'(1) = 2
verify y(x) = x^2 and y(x) = x^-3 are solutions
use wronskian to show both y(x) above are linearly
independent
find unique solution to ivp

Find the first 5 non-zero terms of power series
(1-x^2)y''-2xy'+2y =0

Solve the system by Laplace Transform:
x'=x-2y
y'=5x-y
x(0)=-1, y(0)=2

solve the following initial value problem x' = 2xy
y' = y -2t +1 x(0) = x0 y(0) =
y0

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