# initial value problem for exact equations

#### find_the_fun

##### Active member
I've got a few small questions I'd like to straighten out. I'm really trying to establish a firm procedure involving the steps I write down because I find it helps me learn the math and avoid errors.

Solve the initial value problem: $$\displaystyle (x+y)^2 dx +(2xy+x^2-1)dy = 0$$ with $$\displaystyle y(1)=1$$

So let $$\displaystyle M(x, y) =(x+y)^2$$

Question 1: I write M(x, y) because M is a function of x and y. I that correct or is M only a function of one variable?

Question 2: should dx be part of the equation for M(x, y) or no?

let $$\displaystyle N(x, y) = 2xy+x^2-1$$

$$\displaystyle \frac{\partial M(x, y)}{\partial y}= \frac{\partial}{\partial y} (x^2+y^2+2xy)=2y+2x$$
$$\displaystyle \frac{\partial N(x, y)}{\partial x}=2y+2x$$ therefore equation is exact
$$\displaystyle \frac{ \partial f(x, y)}{\partial x} = M(x, y) = x^2+y^2+2xy$$ implies $$\displaystyle f(x, y)= \int x^2+y^2+2xy dx = \frac{x^3}{3}+y^2x+x^2y+g(y)$$

$$\displaystyle \frac{ \partial f(x, y)}{ \partial y} N(x, y) = 2xy+x^2-1 = 2yx+y+g'(y)$$ solving for $$\displaystyle g'(y)=x^2-y-1$$
$$\displaystyle g(y)= \int x^2-y-1 dy = x^2y-\frac{y^2}{2}-y$$
Question 3: following the examples in my textbook I noticed we don't have a $$\displaystyle C$$ for the constant of integration. Why does it get omitted?
so $$\displaystyle f(x, y) = \frac{x^3}{3}+y^2x+x^2y+x^2y-\frac{y^2}{2}-y=\frac{x^3}{3}+y^2x+2x^2y-\frac{y^2}{2}-y=C$$

Now plugging in initial values

$$\displaystyle \frac{1^3}{3} + 1 +2 - \frac{1^2}{2} -1 = C$$
Question 4: a little embarrassing but does the exponent apply to the whole fraction or just the numerator? For example is $$\displaystyle \frac{1^3}{3}=\frac{1}{3}$$ or $$\displaystyle \frac{1^3}{3} = \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{3}$$? Obviously in LaTex I'm writing it as the 3 only applies to the 1 but given the context that the derivative of $$\displaystyle x^n=(n-1)x^{n-1}$$ which is it?

Question 5: in an example my textbook takes the equation an multiplies away the denominators. Doesn't this mess up the initial values and finding C? For example
$$\displaystyle \frac{1^3}{3} + 1 +2 - \frac{1^2}{2} -1 = C$$
multiply both sides of the equation by 6
$$\displaystyle 2 \cdot 1^3 +6+12-3\cdot 1^2-6=6C_1=C$$

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#### Prove It

##### Well-known member
MHB Math Helper
Where did you get \displaystyle \begin{align*} \tan{(x)} - \sin{(x)}\sin{(y)} \end{align*} from?

#### find_the_fun

##### Active member
Where did you get \displaystyle \begin{align*} \tan{(x)} - \sin{(x)}\sin{(y)} \end{align*} from?
I transcribed the wrong work from the paper to this site.