Question

Solve the IVP y' = 2x(y-1), y(1) = 2

Answer #1

Consider the IVP (x^2 - 2x)dy/dx = (x-2)y + x^2, y(1)=-2. Solve
the IVP. Give the largest interval over which the solution is
defined.

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 IVP: , y(0)=3.
2) Solve the DE: . y' = xy^2/ (x^2 +1)

Solve the initial value problem (IVP): (y + x 2 y)y' = y^2 + 1,
y(0) = 1

1.) 25pt) Solve the IVP: (initial value problem)
y’ = (3x2 + 4x + 2)/(2(y-1)), y(0) = -1

solve the IVP
y'' - 4y' - 5y = 6e-x, y(0)= 1, y'(0) =
-2

Solve the IVP with Cauchy-Euler ODE: x^2 y''+3xy'+4y=0; y(1)=0,
y’(1)=−2

(b) Solve the separable differential
equation
y'=
(7e^-3x+2x^2-xcosx)/-6y^3
(c) Solve the IVP
x(1-siny)dy=(cosx-cosy-y)dx
.
y(π/2)=0

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

Solve the IVP with Cauchy-Euler ODE: x^2 y''+ xy'−16y = 0; y(1)
= 4, y'(1) = 0

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 7 minutes ago

asked 7 minutes ago

asked 12 minutes ago

asked 14 minutes ago

asked 15 minutes ago

asked 15 minutes ago

asked 17 minutes ago

asked 33 minutes ago

asked 54 minutes ago

asked 58 minutes ago

asked 1 hour ago

asked 1 hour ago