# Solve an Inital Value Problem - Possible integrating factor problem?

Question: Solve the initial value problem: $$(y^2+y\sin x\cos y)\,dx+(xy+y\cos x\sin y)\,dy=0$$

So I split up the first term $M=y^2+y\sin x\cos y$ and $N=xy+y\cos x\sin y$. I then determined it was not exact so I used the integrating factor method, in which I got $Y$. Multiplying $Y$ into $M$ and $N$ however still left me with an equation that was not exact, and I've already found 2 integrating factors and still have not gotten exact terms. What could possibly be wrong, or is it not an integrating factor in the first place?

• Please show your work. What did you do and what did you get for the integrating factor? – Dylan Dec 10 '15 at 18:09
• it is $y(x)=0$ one solution and i also got $$-\cos \left( x \right) \cos \left( y \left( x \right) \right) +xy \left( x \right) +{\it _C1}=0$$ – Dr. Sonnhard Graubner Dec 10 '15 at 18:14

The first part of your solution is correct, including $y$ as the overall multiplying factor for the equation. So, to summarise:
The equation is of the form $M(x,y) + N(x,y) \frac{\text{d} y}{\text{d} x} = 0$ (note that I don't separate $\text{d}y$ and $\text{d} x$, for good reason). Initially, the hope is that this equation is exact, that is, that it can be written as the total derivative to $x$ of an unknown function $H(x,y(x))$, which looks like $$\frac{\text{d}}{\text{d} x} H(x,y(x)) = H_x(x,y(x)) + H_y(x,y(x)) \frac{\text{d} y}{\text{d} x}.$$ So, if your equation would be exact, we would have $M(x,y) = H_x(x,y)$ and $N(x,y) = H_y(x,y)$. In particular, this means that $M_y(x,y) = H_{xy}(x,y) = N_x(x,y)$. Unfortunately, this is not the case, as you found yourself.
The next attempt is to try to find a common factor in $M$ and $N$, i.e. assume that our equation is of the form
$$Y(x,y) H_x(x,y) + Y(x,y) H_y(x,y) \frac{\text{d} y}{\text{d} x} = 0,$$ which means that $M = Y H_x$ and $N = Y H_y$. Performing the same differentiation trick as above, we see that then it must be true that $$M_y = Y_y H_x + Y H_{xy} \quad \text{and} \quad N_x = Y_x H_y + Y H_{xy},$$ so $$M_y - N_x = Y_y H_x - Y_x H_y = \frac{Y_y}{Y} M - \frac{Y_x}{Y} N.$$
As you calculated, the difference $M_y - N_x$ is in our case $$M_y - N_x = y + \cos y \sin x,$$ which we recognise as $\frac{M}{y}$! That means that $Y$ is a function of $y$ only, and that $$\frac{Y_y}{Y} = \frac{1}{y}.$$ It is easy to solve this equation, and we obtain $Y(y) = y$, as you did too in your calculations.
Now let's reflect on the above. We assumed that $M$ and $N$ were of the form $M = Y H_x$ and $N = Y H_y$. We've found $Y$, so we can use that to find $H_x$ and $H_y$, since $$H_x = \frac{M}{Y} \quad \text{and}\quad H_y = \frac{N}{Y}.$$ By integrating the first expression with respect to $x$ or the second with respect to $y$, we find $H(x,y)$. This is where you made a mistake: instead of dividing $M$ and $N$ by $Y$, you multiplied by $Y$.