Exact Differential Equations: (3x^2 y^2 - 3y^2) dx + (2x^3y - 6xy + 3y^2) dy =...

Exact Differential Equations:

(3x^2 y^2 - 3y^2) dx + (2x^3y - 6xy + 3y^2) dy = 0

Expert Solution

Colby Messinger answered 1 year ago

Integrate the following differential equations: a.) y(2x−y+2)dx+2(x−y)dy = 0; b.) yV-2yIV+3y''' - 2y'' = -x + ex.

Solve ODE (3x - 2y + 1)dx + (-2x + y + 2)dy = 0 with...

Solve ODE (3x - 2y + 1)dx + (-2x + y + 2)dy = 0 with the method of x =u + h and y = v + k

Initial value problem : Differential equations: dx/dt = x + 2y dy/dt = 2x + y...

Initial value problem : Differential equations: dx/dt = x + 2y dy/dt = 2x + y Initial conditions: x(0) = 0 y(0) = 2 a) Find the solution to this initial value problem (yes, I know, the text says that the solutions are x(t)= e^3t - e^-t and y(x) = e^3t + e^-t and but I want you to derive these solutions yourself using one of the methods we studied in chapter 4) Work this part out on paper to...

Let Q1=y(2) Q2 =y(3) where y=y(x) solves y'+2xy=2x^3 y(0)=1 dy/dx= (2x-y)/(x+4y) y(1)=1 y'+ycotx=y^3sin^3x y(pi/2)=1 (y^2-2x)dx+2xy dy=0 ...

Let Q1=y(2) Q2 =y(3) where y=y(x) solves y'+2xy=2x^3 y(0)=1 dy/dx= (2x-y)/(x+4y) y(1)=1 y'+ycotx=y^3sin^3x y(pi/2)=1 (y^2-2x)dx+2xy dy=0 y(1)=1 2x^2y;+4xy=3sinx y(2pi)=0 y"+2y^-1 (y')^2=y'

(2x−y3−yex)dx=(2+ex+3xy2 )dy, y(0)=2. Solve the initial value problem and verify exact.

find the differential equation of dy/dx = y+y^3

Question

Exact Differential Equations: (3x^2 y^2 - 3y^2) dx + (2x^3y - 6xy + 3y^2) dy =...

Solutions

Expert Solution

Related Solutions

Integrate the following differential equations: a.) y(2x−y+2)dx+2(x−y)dy = 0; b.) yV-2yIV+3y''' - 2y'' = -x + ex.

Solve ODE (3x - 2y + 1)dx + (-2x + y + 2)dy = 0 with...

Initial value problem : Differential equations: dx/dt = x + 2y dy/dt = 2x + y...

Solve the following differential equations. a.) (2xy^2 +2x)dx−(4x^2 +1)dy=0 b.) (3ye^(3xy) +4xy)dx+(3xe^(3xy) +2x^2)dy=0

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

Find the general solution for differential equation x^3y'''-(3x^2)y''+6xy'-6y=0, y(1)=2, y'(1)=1, y''(1)=-4

Use the Laplace transform to solve the given system of differential equations. dx/dt + 2x +dy/dt=...

Let Q1=y(2) Q2 =y(3) where y=y(x) solves y'+2xy=2x^3 y(0)=1 dy/dx= (2x-y)/(x+4y) y(1)=1 y'+ycotx=y^3sin^3x y(pi/2)=1 (y^2-2x)dx+2xy dy=0 ...

(2x−y3−yex)dx=(2+ex+3xy2 )dy, y(0)=2. Solve the initial value problem and verify exact.

find the differential equation of dy/dx = y+y^3