Differential Equations Help!

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.
$$e^{\lambda x} + e^{\lambda x} \lambda = e^{\lambda x} \lambda^2$$

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saltydog

Homework Helper
Orion1 said:
Find an equation of the tangent line to the curve:
$$xe^y + ye^x = 1$$
at the point:
$$P(0,1)$$
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.

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

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

This is my first attempt at problem 2, uncertain if this is correct.
$$e^{\lambda x} + e^{\lambda x} \lambda = e^{\lambda x} \lambda^2$$
[/COLOR]
Looks good. You can figure out what lambda is right?

dextercioby

Homework Helper
The theorem of implicit functions solves the first problem.

Daniel.

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:

dextercioby

Homework Helper
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.

Orion1

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

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

Last edited:

dextercioby

Homework Helper
Yes,it looks good now.

Daniel.

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