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#### find_the_fun

##### Active member

- Feb 1, 2012

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

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 \(\displaystyle g'(y)=x^2-y-1\)

\(\displaystyle g(y)= \int x^2-y-1 dy = x^2y-\frac{y^2}{2}-y\)

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

\(\displaystyle \frac{1^3}{3} + 1 +2 - \frac{1^2}{2} -1 = C\)

\(\displaystyle \frac{1^3}{3} + 1 +2 - \frac{1^2}{2} -1 = C\)

\(\displaystyle 2 \cdot 1^3 +6+12-3\cdot 1^2-6=6C_1=C\)

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?

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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