Question

Solve the separable equation 3xy^2 dy/dx = x^2 + 1 given y(0) =5

I am getting a general solution of *cubed root
x ^{2}/2+ln|x|+3c_{1}* when I enter the initial
conditions y(0)=5 the equation is undefined. How do deal with this?
What would the final answer be?

Thanks

Answer #1

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 following differential equation: dy/dx =
−(3xy+y^2)/x^2+xy
(a) Rewrite this equation into the form M(x, y)dx + N(x, y)dy =
0. Determine if this equation is exact;
(b) Multiply x on both sides of the equation, is the new
equation exact?
(c) Solve the equation based on Part (a) and Part (b).

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

1) Solve the given differential equation by separation of
variables.
exy
dy/dx = e−y +
e−6x −
y
2) Solve the given differential
equation by separation of variables.
y ln(x) dx/dy = (y+1/x)^2
3) Find an explicit solution of the given initial-value
problem.
dx/dt = 7(x2 + 1), x( π/4)= 1

Let F(x,y) = x^2 +3xy+4y^2 −14. Solve the equation x^2 +3xy+4y^2
−14 = 0
by the quadratic formula for x in terms of y. Determine dx/dy when
y = 1.
Find the function y = f(x) for which F(x,f(x)) = 0 and f(2) = 1.
Determine f′(2). How is f′(2) related to the value of dx/dy that
you found above.

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!

(x-y)dx + (y+x)dy =0 Solve the differential equation

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 given initial-value problem. (x + 2) dy dx + y =
ln(x), y(1) = 10 y(x) =
Give the largest interval I over which the solution is defined.
(Enter your answer using interval notation.)
I =

Solve the Homogeneous differential equation
(7 y^2 + 1 xy)dx - 1 x^2 dy = 0
(a) A one-parameter family of solution of the equation is y(x)
=
(b) The particular solution of the equation subject to the
initial condition y(1) =1/7.

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