# Differential Equations Help!

1. Jun 16, 2005

### Orion1

Find an equation of the tangent line to the curve:
$$xe^y + ye^x = 1$$
at the point:
$$P(0,1)$$

Find the values for $$\lambda$$ for which:
$$y = e^{\lambda x}$$

satisfies the equation:
$$y + y' = y''$$

I have been assigned these two problems, which are not covered for another 3 chapters.

I am uncertain how to solve problem 1.

This is my first attempt at problem 2, uncertain if this is correct.
$$e^{\lambda x} + e^{\lambda x} \lambda = e^{\lambda x} \lambda^2$$

2. Jun 16, 2005

### saltydog

Differentiate implicitly throughout the expression. You can do that right? You know, when you do so for:

$$xe^y[/itex] you get [tex] xe^y(dy)+e^y(dx)$$

Right? Now do the rest, then solve for the quotient dy/dx.

Looks good. You can figure out what lambda is right?

3. Jun 16, 2005

### dextercioby

The theorem of implicit functions solves the first problem.

Daniel.

4. Jun 16, 2005

### Orion1

Implicit Function Theorem...

The correct application of the Implicit Function Theorem:

$$\frac{dy}{dx} = - \frac{F_x}{F_y} = - \frac{e^y + ye^x}{xe^y + e^x}$$

Solution 1:
$$\boxed{\frac{dy}{dx} = - \frac{e^y + ye^x}{xe^y + e^x}}$$

Solution 2:
$$y = e^{\lambda x} \; \; y + y' = y''$$
$$e^{\lambda x} + e^{\lambda x} \lambda = e^{\lambda x} \lambda^2$$
$$e^{\lambda x} (1 + \lambda) = e^{\lambda x} \lambda^2$$
$$(1 + \lambda) = \lambda^2$$
$$\lambda^2 - \lambda = 1$$
$$\lambda^2 - \lambda - 1 = 0$$
$$\lambda = \frac{1 \pm \sqrt{(-1)^2 - 4(1)(-1)}}{2(1)} = \frac{1 \pm \sqrt{5}}{2}$$
$$\boxed{\lambda = \frac{1 \pm \sqrt{5}}{2}}$$

Last edited: Jun 16, 2005
5. Jun 16, 2005

### dextercioby

Nope

$$\frac{dy}{dx}=-\frac{e^{y}+ye^{x}}{xe^{y}+e^{x}}$$

computed at the point (0,1) gives the slope $-e-1$.

Daniel.

6. Jun 16, 2005

### Orion1

Slope formula:
$$(y - 1) = (-e - 1)(x - 0)$$

Solution 1:
$$\boxed{y(x) = -x(e + 1) + 1}$$

Last edited: Jun 16, 2005
7. Jun 16, 2005

### dextercioby

Yes,it looks good now.

Daniel.