What Are the Solutions for These Two Differential Equations?

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

 

[saltydog]

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:

[tex]xe^y[/itex]<br /> <br /> you get<br /> $$xe^y(dy)+e^y(dx)$$<br /> <br /> Right? Now do the rest, then solve for the quotient dy/dx.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Find the values for $$\lambda$$ for which:<br /> $$y = e^{\lambda x}$$<br /> <br /> satisfies the equation:<br /> $$y + y' = y''$$<br /> <br /> This is my first attempt at problem 2, uncertain if this is correct.<br /> $$e^{\lambda x} + e^{\lambda x} \lambda = e^{\lambda x} \lambda^2$$ </div> </div> </blockquote><br /> Looks good. You can figure out what lambda is right?[/tex]

 

[dextercioby]

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

 

[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.

 

[Orion1]

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

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

 

[dextercioby]

Yes,it looks good now.

Daniel.