Solutions to Differential equations

[Locoism]

I'm taking my first ODE course, and I'm unsure what is meant in this question when it asks "Solve the following differential equation" (I have a list of DEs to solve). Some of them are really messy and I can't figure out an implicit solution. Would an explicit solution be acceptable?

Example:

\frac{dy}{dx} = \frac{x}{x^2y+y^3}

Since this isn't an exact equation, I transform it and find the integrating factor

\frac{dy}{dx}(x^2y+y^3) - x = 0

and setting the partials \frac{∂}{∂x}N(x,y)μ(x) = \frac{∂}{∂y}M(x,y)μ(x)

I get μ(x) = \frac{1}{x^2+y^2}

I thought the integrating factor was supposed to be only a function of x, I may have made a mistake there.

Then I take \frac{∂}{∂y}\int M(x,y)μ(x) dx

and I find f(x,y)=C=\frac{x^2+y^2}{2} - ln(x^2+y^2)


This just doesn't look right, I feel like I'm using the wrong method. Can someone clear this up for me please?

 

[jackmell]

I'm taking my first ODE course, and I'm unsure what is meant in this question when it asks "Solve the following differential equation" (I have a list of DEs to solve). Some of them are really messy and I can't figure out an implicit solution. Would an explicit solution be acceptable?

Example:

\frac{dy}{dx} = \frac{x}{x^2y+y^3}

Since this isn't an exact equation, I transform it and find the integrating factor

\frac{dy}{dx}(x^2y+y^3) - x = 0

and setting the partials \frac{∂}{∂x}N(x,y)μ(x) = \frac{∂}{∂y}M(x,y)μ(x)

I get μ(x) = \frac{1}{x^2+y^2}

I thought the integrating factor was supposed to be only a function of x, I may have made a mistake there.

Then I take \frac{∂}{∂y}\int M(x,y)μ(x) dx

and I find f(x,y)=C=\frac{x^2+y^2}{2} - ln(x^2+y^2)


This just doesn't look right, I feel like I'm using the wrong method. Can someone clear this up for me please?

First convert it to:

Mdx+Ndy=0

You can do that. I get:

xdx-(x^2 y+y^3)dy=0

Now compute:

\mu=\frac{1}{M}\left(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}\right)

I get \mu=2y but I did that quick so double-check it. Therefore, the integrating factor is:

e^{-\int \mu dy}

Now apply that integrating factor to the DE and solve it as an exact DE. And to "solve" a DE is to find a function y(x) for example, which when substituted into the DE, satisfies the expression.

 

[Locoism]

Yes that's right, but I was under the impression the integrating factor would be a function of x, seeing as when we assume ∂μ/∂y = 0, this would be false if μ = 2y, and consequently fail to make the equation exact. So can the integrating factor be a function of y, or even both x and y?

 

[jackmell]

Yes that's right, but I was under the impression the integrating factor would be a function of x, seeing as when we assume ∂μ/∂y = 0, this would be false if μ = 2y, and consequently fail to make the equation exact. So can the integrating factor be a function of y, or even both x and y?

Yes, integrating factor can be a function of x and y for example:

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

the integrating factor is \frac{1}{yx^2}

Find "Differential Equations" by Rainville and Bedient. Good book I think.

 

[HallsofIvy]

Generally speaking, "implicit" solutions cannot be written as explicit functions of x. Whether your instructor will accept such solutions even when, by trying a little harder, you could have found an explicit function, you will have to ask your instructor!