# Homework Help: Partial Derivatives

1. Oct 2, 2009

### KillerZ

1. The problem statement, all variables and given/known data

Determine if the following differential equation is exact. If it is exact solve it.

2. Relevant equations

$$\left(\frac{1}{t} + \frac{1}{t^{2}} - \frac{y}{t^{2} + y^{2}}\right)dt + \left(ye^{y} + \frac{t}{t^{2} + y^{2}}\right)dy = 0$$

3. The attempt at a solution

I am a little rusty on my partial derivatives I am not sure if this is right.

$$M(t, y) = \frac{1}{t} + \frac{1}{t^{2}} - \frac{y}{t^{2} + y^{2}}$$

$$N(t, y) = ye^{y} + \frac{t}{t^{2} + y^{2}}$$

$$\frac{\partial M}{\partial y} = -y[-(t^{2} + y^{2})^{-2}(2y)] - (t^{2} + y^{2})^{-1}$$

$$\frac{\partial N}{\partial t} = t[-(t^{2} + y^{2})^{-2}(2t)] + (t^{2} + y^{2})^{-1}$$

2. Oct 2, 2009

### Dick

Looks good so far. You'll have to do some simplifying to see if they are the same. Put them over a common denominator.

3. Oct 2, 2009

### KillerZ

$$-y[-(t^{2} + y^{2})^{-2}(2y)] - (t^{2} + y^{2})^{-1} = t[-(t^{2} + y^{2})^{-2}(2t)] + (t^{2} + y^{2})^{-1}$$

$$-y\left[-\frac{2y}{(t^{2} + y^{2})^{2}}\right] - \frac{1}{t^{2} + y^{2}} = t\left[-\frac{2t}{(t^{2} + y^{2})^{2}}\right] + \frac{1}{t^{2} + y^{2}}$$

$$\frac{2y^{2}}{(t^{2} + y^{2})^{2}} + \frac{2t^{2}}{(t^{2} + y^{2})^{2}} = \frac{1}{t^{2} + y^{2}} + \frac{1}{t^{2} + y^{2}}$$

$$\frac{2y^{2} + 2t^{2}}{(t^{2} + y^{2})^{2}} = \frac{2}{t^{2} + y^{2}}$$

$$\frac{2y^{2} + 2t^{2}}{t^{2} + y^{2}} = 2$$

$$2y^{2} + 2t^{2} = 2y^{2} + 2t^{2}$$ They are exact.

Last edited: Oct 2, 2009
4. Oct 2, 2009

### KillerZ

To finish it:

$$\frac{\partial f}{\partial t} = \frac{1}{t} + \frac{1}{t^{2}} - \frac{y}{t^{2} + y^{2}}$$

$$\frac{\partial f}{\partial y} = ye^{y} + \frac{t}{t^{2} + y^{2}}$$

$$f(t,y) = \int\left(\frac{1}{t} + \frac{1}{t^{2}} - \frac{y}{t^{2} + y^{2}}\right)dt + \Phi(y)$$

$$f(t,y) = ln|t| - \frac{1}{t} - tan^{-1}\left(\frac{t}{y}\right) + \Phi(y)$$

$$\frac{\partial f}{\partial y} = 0 - 0 + \frac{t}{y^{2} + t^{2}} + \frac{d\Phi}{dy} = ye^{y} + \frac{t}{t^{2} + y^{2}}$$

$$\frac{d\Phi}{dy} = ye^{y}$$

$$\Phi = \int\left(ye^{y}\right)dy + c$$

$$\Phi = ye^{y} - e^{y} + c$$

$$f(t,y) = ln|t| - \frac{1}{t} - tan^{-1}\left(\frac{t}{y}\right) + ye^{y} - e^{y} + c$$

5. Oct 2, 2009

### Dick

Sure. Well done!

6. Oct 2, 2009

### KillerZ

Thanks for the help.