Is There More Than One Method to Solve Nonhomogeneous Differential Equations?

[Leptos]

Homework Statement

y-(sinx)²)dx + sinx)dy

Homework Equations

Since the result wouldn't be a line, the equation would only be linear in one of its variables.

The Attempt at a Solution

y-(sinx)² = 0 ; sinx = 0 --- y = (sinx)² + sinx

No clue... Also, is there more than one way to solve this? For example I came across a couple problems that could be solved by as many as 3 different techniques.

I'm wondering how I can get a general feel for these problems because it's not always immediately obvious what approach should be used to solve them(of course that's the point of practicing)... For example when should I generally try determining the integration factor: M_y - N_x/M = f(x) where u = e^int(f(x)) is the I.F. Or things like reducing a problem into common differentials such as d(xy), d(y/x), d(x³/y) or making several changes of variables... The last one is really troublesome for me because I find that picking one way say ZX = V can take one order of magnitude longer than picking ZV = X in some cases, it's hard to develop this intuition of which method is better, or which works, etc. Any advice?

 

[Stephen Tashi]

First let's try to write an equation instead of an expression. Is it supposed to be:

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

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

 

[Leptos]

First let's try to write an equation instead of an expression. Is it supposed to be:

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

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

Yes... I know that I have to use the method of linear coefficients but I'm not sure how to go about using it in this case.

 

[Stephen Tashi]

I don't know what "the method linear cofficients" is. Do you have a link to it?

Are you sure the equation isn't:

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

 

[Leptos]

I don't know what "the method linear cofficients" is. Do you have a link to it?

Are you sure the equation isn't:

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

That's the same as the first line in the OP.
Yes, ydx - (sinx)²dx +sinxdy = 0

We would treat the two parts under dx and dy separately as follows:
E₁ = _______dx
E₂ = _______dy
And we would try to find where E₁ and E₂ intersect. In the case they don't intersect, they must be parallel therefore there will be some part that repeats so in that case it's just substitution.

For example: (x + 2y - 4)dx - (2x + y - 5)dy = 0
E₁ is x + 2y - 4 = 0
E₂ is 2x + y - 5 = 0
We can easily see they are not parallel(to be parallel E₂ would have to be x + 2y + constant = 0) and so we solve the system of equations and find an intersection at (2,1) meaning we want a change of variables that will transform the equation so the origin is shifted to (h,k) = (2, 1)

Thus we change variables: u = x + h = x + 2 and v = y + k = y + 1
Thus the original equation is transformed to (u + 2v)du - (2u + v)dv and from here we would make a second change of variables since the equation in u and v is homogenous and of degree 1 in said variables. We get u = vz and du = vdz + zdv which transforms the equation into something separable...

That being said, I have no clue how to handle the equation I posted in the OP.

 

[snipez90]

Generally if an equation is of that form, I do check if it's exact since it takes a second. If the equation is not exact, as in this case, I probably rewrite the equation in the form y' = f(x,y) to see if I can work with that.

Doing this gives y' = sin(x) - y/sin(x), which is a first-order nonhomgeneous equation. You should have known how to solve this before you encountered exact equations.

Also you can make this equation exact simply by multiplying the original equation through by the integrating factor p, imposing the conditions for exactness, then setting p = p(x) (stipulating that p is a function of x alone) and solving a separate first order equation.

However, there is no reason to do this if you realized that your equation is actually of first order in the first place.