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.