Note the LHS is d/dx (y/x).Lol they made us solve this differential equation by considering how it is an exact ODE. But the solution y=Ax just bugs me... is there a shorter way to doing this one haha
Oops..........................
Reverse product?Oops..........................
That moment when I saw it for this harder one but not the easier one
Yeah, I'm saying I missed reverse product for the first but saw it for the e^(xy) one
Final answer ends up with exp(xy)cos(x)=C though
Remember that d/dz z^n = nz^(n-1), I think that should help you...This is boring and tiring. What is the giveaway as to what values of p and q I am supposed to pick.
dx+(4xy+3x^3)dy=0\\$ can be turned into an exact equation by multiplying throughout by a function of the form $x^py^q$, where $p$ and $q$ are constants. Hence solve the equation.$)
It's hard to juggle that when I have to consider two constants at a time and not just one
This is boring and tiring. What is the giveaway as to what values of p and q I am supposed to pick.
y is on the right, and there is a 3, so you want 3y². x is on the left and you have 4x² so you want 4x³. so xy² is a possibility...
Andddd this is where intuition collides head on with mechanical rigour...
x^p y^{1+q} + 4(1+q)x^{2+p} y^q$ and $\frac{\partial B}{\partial x} = 4(1+p)x^{p} y^{1+q} + 3(3+p) x^{2+p}y^q$. We want these two things to be equal for the ODE to be exact, so equating coefficients of like terms, we clearly will take $2(2+q) = 4(1+p)$ and $4(1+q) = 3(3+p)$. Solve this system of simultaneous equations and we'll have the required values of $p$ and $q$. Once we have these values, the rest is just the usual tedium.$)
make dy/dx the subject and you'll see.I can't identify the type of this one...
...unless I did my partial derivatives wrong
Oh crap I forgot about that g(y/x) because they said they'd give us the substitution in the exam
Once you apply the substitution, maybe you could differentiate both sides again?
Part a) was quite ok
Part b)... what? How does that even work