Toronto Math Forum
MAT2442020F => MAT244Lectures & Home Assignments => Chapter 2 => Topic started by: Julian on September 28, 2020, 12:45:14 PM

In week 3 lecture 3, we get the example $(y\sin(x)+y^3\cos(x))dx+(3\cos(x)+5y^2\sin(x))dy=0$. We determine that this equation is not exact, but that we can make it exact by multiplying the equation by $y^2$. We then find the general solution to the new equation is $y^3\cos(x)+y^5\sin(x)=C$.
My question is why is this good enough? We didn't answer the original question. We answered a modified version which we chose specifically because it seems easier to us. Shouldn't we still find a solution to the original equation $(y\sin(x)+y^3\cos(x))dx+(3\cos(x)+5y^2\sin(x))dy=0$? Is there some way we can "divide out" $y^2$ from $y^3\cos(x)+y^5\sin(x)=C$ to get it?

In week 3 lecture 3, we get the example $(y\sin(x)+y^3\cos(x))dx+(3\cos(x)+5y^2\sin(x))dy=0$. We determine that this equation is not exact, but that we can make it exact by multiplying the equation by $y^2$. We then find the general solution to the new equation is $y^3\cos(x)+y^5\sin(x)=C$.
My question is why is this good enough? We didn't answer the original question. We answered a modified version which we chose specifically because it seems easier to us. Shouldn't we still find a solution to the original equation $(y\sin(x)+y^3\cos(x))dx+(3\cos(x)+5y^2\sin(x))dy=0$? Is there some way we can "divide out" $y^2$ from $y^3\cos(x)+y^5\sin(x)=C$ to get it?
We have not modified equation, but simply multiplied it by an integrating factor. These two equations are equivalent, except as $y=0$, that means $C=0$. But $y=0$ os also a solution to the original equation. Checking this would give you a 100% correct solution, otherwise it is almost perfect