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:

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&amp;#039; = y&amp;#039;&amp;#039;<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?

 

[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&#039; = y&#039;&#039;
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.