Differential Equation help :(

Here's the question (click to enlarge):



Tried to separate the variables after using some trig identities but found I ended up with a sinx on the dy side and visa versa. Pretty new to differential equations, not sure where to go from here? And when to introduce an integrating factor? And if so how do you know what it'll be?

Thanks a lot

If you can't seperate the variables and you have an equation of the form dy/dx +Py = Q, then it looks like you should find the integrating factor. Do you know how to find it from that equation?

And if you have sinx dy = sin y dx, or anything like this, think of how you could move the siny to the dy side and vice versa.

I thought the integrating factor method only worked for a linear differential equation? This is certainly not linear since we have siny and cosy terms!

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To solve this you need to employ multivariable calculus, and partial differentiation, which isn't covered at A-Level, so this is probably a degree level question.

Imagine you have a non-linear differential equation such as the one shown in the OP. The general form for a non-linear first order differential equation is: $M(x,y) + N(x,y) \dfrac{dy}{dx}$.

If you rearrange the equation given so that it is in this form you end up with $(-cosxcosy - sin^{2}x) + (sinxsiny + cos^{2}y) \dfrac{dy}{dx}$.

Now for a differential equation with an exact solution, imagine that there is a function $F(x,y)$; taking the partial derivative of with respect to each variable gives us the functions $F_{x}$ and $F_{y}$, if this function is such that $F_{x}$ and $F_{y}$ are equal to $M(x,y)$ and $N(x,y)$ respectively, then we can solve the equation to find $F(x,y)$.

If you employ the logic that $M(x,y)$ is the partial derivative of $F(x,y)$ taken with respect to x, and $N(x,y)$ is the equivalent taken with respect to y. If there is an exact solution for $F(x,y)$, then if you take the partial derivative of $M(x,y)$ with respect to y, it should be equal to the partial derivative of$N(x,y)$ taken with respect to x, and in fact that is a quick way to test if there is an exact solution.

$M(x,y) = -cosxcosy - sin^{2}x \therefore M_{y} = cosxsiny$

$N(x,y) = sinxsiny + cos^{2}y \therefore N_{x} = cosxsiny$

It is clear that the two partial derivatives are equal, consequently there exists a function $F(x,y)$ which is the solution to this differential equation.

We know that $M(x,y)$ is equal to $F_{x}$, therefore if we integrate $M(x,y)$ with respect to x we should end up with $F(x,y)$.

$\int -cosxcosy - sin^{2}x dx = -sinxcosy -\dfrac{x}{2} + \dfrac{sin2x}{4} + f(y)$

Note the addition of a function of y at the end of this integral; we are dealing with partial derivatives so the typical integration constant is replaced by a function in the variable we didn't integrate with.

So now we know what $F(x,y)$ is equal to, but we still have that function in y at the end. We also know that $F_{y}$ is equal to $N(x,y)$, therefore:

$\dfrac{\partial}{\partial y}[-sinxcosy - \dfrac{x}{2} + \dfrac{sin2x}{4} + f(y)] = sinxsiny + f'(y)$

$sinxsiny + f'(y) = sinxsiny + cos^{2}y$

$\therefore f'(y) = cos^{2}y$

$\therefore f(y) = \dfrac{y}{2} + \dfrac{sin2y}{4}$

$\therefore F(x,y) = -sinxcosy -\dfrac{x}{2} + \dfrac{sin2x}{4} + \dfrac{y}{2} + \dfrac{sin2y}{4}$

And that is the solution to your differential equation. I have given a full solution here as if you haven't worked with partial derivatives before you likely won't know what to do with this type of question. I hope this helped you!

Where did you learn this?

totally agreed! (I have no paper and try to picture and I think I get a different sign somewhere but you are probably correct)

Where did you learn this?

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Sometimes I'm a bit sad and just read into maths on the internet A good website for this stuff is http://tutorial.math.lamar.edu/Classes/DE/Exact.aspx - the other pages on the website are brilliant as well!

look at this booklet at the very end when it deals with applications of partial differentiation to ODEs
http://madasmaths.com/archive/maths_booklets/advanced_topics/partial_differentiation.pdf

Just looked in your profile
Truly amazed!
Not completed A levels in Maths and can do this type of ODE so competently.
Keep it up

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Haha thank you mate! Maths is just something I really enjoy reading into so it means a lot that someone thinks that is a good thing

No worries

Rarely I meet people that are just interested.
the usual is people that are just interested how to pass exams, or how to become" bankers".

Fantastic stuff.

(will rep +5 your excellent explanation tonight when the system lets me)

No worries

Rarely I meet people that are just interested.
the usual is people that are just interested how to pass exams, or how to become" bankers".

Fantastic stuff.

(will rep +5 your excellent explanation tonight when the system lets me)

No worries

Rarely I meet people that are just interested.
the usual is people that are just interested how to pass exams, or how to become" bankers".

Fantastic stuff.

(will rep +5 your excellent explanation tonight when the system lets me)

careful not to get a hard on mate

Interestingly enough, I've just been told in another thread that I'm wasting my time going to university since I go simply to learn.



I find it's impossible not to get one when I'm talking/learning about differential equations. So awkward in my differential equation lectures.

Yh but shes getting one abput the boy not DE. DEs probably my easiestcourse this year so its not very exciting

Obviously she's not because women can't get them. Each to their own I suppose.