Question

Solve:

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

Answer #1

solve dy/dx + y/x = 8e-2x

Solve: 1.dy/dx=(e^(y-x)).secy.(1+x^2),y(0)=0.
2.dy/dx=(1-x-y)/(x+y),y(0)=2 .

Solve the following differential equations.
a.) dy/dx+2xy=x, y(0)=2
b.) ?^2(dy/dx)−?y=−y^2

Solve the equations: dy/dx+(1/x-2x/(1-x^2))*y=1/(1-x^2)

Solve the differential equation (5x^4 y^2+ 2xe^y - 2x cos (x^2)) dx
+ (2x^5y + x^2 e^y) dy = 0.

Solve the equation.
(2x^3+xy)dx+(x^3y^3-x^2)dy=0
give answer in form F(x,y)=c

solve diffeential equation.
( x2y +xy -y )dx + (x2 y -2 x2)
dy =0 answer x + ln x + x-1 + y- 2 lny = c
dy / dx + 2y = e-2x - x^2 y (0) =3 answer
y = 3 e -2s + e-2x ( intefral of e
-s^2ds ) s is power ^ 2 means s to power of
2

solve for y:
1. dy/dx= xe^y separable
2. x(dy/dx)+3y=x^2 when y(5)=0 1st order linear

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 (2xy/x^2+1-2x)dx-(2-ln(x^2+1))dy=0, y(5)=0. Write the
solution in the explicit form. please explain steps Thanks!

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