Differential Equations Help!

  • Thread starter Orion1
  • Start date
969
2
Find an equation of the tangent line to the curve:
[tex]xe^y + ye^x = 1[/tex]
at the point:
[tex]P(0,1)[/tex]

Find the values for [tex]\lambda[/tex] for which:
[tex]y = e^{\lambda x}[/tex]

satisfies the equation:
[tex]y + y' = y''[/tex]


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.
[tex]e^{\lambda x} + e^{\lambda x} \lambda = e^{\lambda x} \lambda^2[/tex]
 

saltydog

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Orion1 said:
Find an equation of the tangent line to the curve:
[tex]xe^y + ye^x = 1[/tex]
at the point:
[tex]P(0,1)[/tex]
Differentiate implicitly throughout the expression. You can do that right? You know, when you do so for:

[tex]xe^y[/itex]

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

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

Find the values for [tex]\lambda[/tex] for which:
[tex]y = e^{\lambda x}[/tex]

satisfies the equation:
[tex]y + y' = y''[/tex]

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

dextercioby

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The theorem of implicit functions solves the first problem.

Daniel.
 
969
2
Implicit Function Theorem...



The correct application of the Implicit Function Theorem:

[tex]\frac{dy}{dx} = - \frac{F_x}{F_y} = - \frac{e^y + ye^x}{xe^y + e^x}[/tex]

Solution 1:
[tex]\boxed{\frac{dy}{dx} = - \frac{e^y + ye^x}{xe^y + e^x}}[/tex]

Solution 2:
[tex]y = e^{\lambda x} \; \; y + y' = y''[/tex]
[tex]e^{\lambda x} + e^{\lambda x} \lambda = e^{\lambda x} \lambda^2[/tex]
[tex]e^{\lambda x} (1 + \lambda) = e^{\lambda x} \lambda^2[/tex]
[tex](1 + \lambda) = \lambda^2[/tex]
[tex]\lambda^2 - \lambda = 1[/tex]
[tex]\lambda^2 - \lambda - 1 = 0[/tex]
[tex]\lambda = \frac{1 \pm \sqrt{(-1)^2 - 4(1)(-1)}}{2(1)} = \frac{1 \pm \sqrt{5}}{2}[/tex]
[tex]\boxed{\lambda = \frac{1 \pm \sqrt{5}}{2}}[/tex]
 
Last edited:

dextercioby

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Nope

[tex] \frac{dy}{dx}=-\frac{e^{y}+ye^{x}}{xe^{y}+e^{x}} [/tex]

computed at the point (0,1) gives the slope [itex] -e-1 [/itex].

Daniel.
 
969
2

Slope formula:
[tex](y - 1) = (-e - 1)(x - 0)[/tex]

Solution 1:
[tex]\boxed{y(x) = -x(e + 1) + 1}[/tex]
 
Last edited:

dextercioby

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Yes,it looks good now.

Daniel.
 

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