Orthogonal Trajectory. Calc IV.

In summary, the problem is finding orthogonal trajectories for three given equations and utilizing simple integration techniques to solve them. The first two equations are relatively straightforward, but the third one is causing difficulty. The attempt at a solution involves finding the value of a constant, c, and using it to find the slope of the orthogonal trajectories. However, the integration process becomes complex and further assistance is needed to progress.
  • #1
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Homework Statement



Find orthogonal trajectories for y=clnx, y=cex, y=sin(x)+cx

Homework Equations



Simple integration for the most part.

The Attempt at a Solution



I'm fairly confident on the first two, its the third that's giving me trouble.

First, y=cln(x)
c=[itex]\frac{y}{ln(x)}[/itex]
[itex]\frac{dy}{dx}[/itex]=[itex]\frac{c}{x}[/itex]
[itex]\frac{dy}{dx}[/itex]=[itex]\frac{y}{xln(x)}[/itex]=m1
[itex]\frac{dy}{dx}[/itex]=[itex]\frac{-xln(x)}{y}[/itex]=m2
[itex]\int[/itex]-xln(x)dx=[itex]\int[/itex]ydy using IBP
[itex]\frac{-lnxx^2}{2}[/itex]-[itex]\int[/itex][itex]\frac{-x}{2}[/itex]dx
c+[itex]\frac{-lnxx^2}{2}[/itex]+[itex]\frac{x^2}{4}[/itex]=[itex]\frac{y^2}{2}[/itex]

Second, y=cex
c=[itex]\frac{y}{e^x}[/itex]
[itex]\frac{dy}{dx}[/itex]=cex
[itex]\frac{dy}{dx}[/itex]=[itex]\frac{e^xy}{e^x}[/itex]=y=m1
[itex]\frac{dy}{dx}[/itex]=[itex]\frac{-1}{y}[/itex]=m2
[itex]\int[/itex]ydy=[itex]\int[/itex]-1dx
[itex]\frac{y^2}{2}[/itex]+x=c

Third, y=sin(x)+cx2
c=[itex]\frac{y-sin(x)}{x^2}[/itex]
[itex]\frac{dy}{dx}[/itex]=cos(x)+2xc
[itex]\frac{dy}{dx}[/itex]=cos(x)+[itex]\frac{2(y-sin(x))}{x}[/itex]

I've been unable to make any more progress.
 
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  • #2
Can nobody give me an idea of where I should go from here? I don't know if I'm having a simple algebra problem or what.
 
  • #3
Why won't anyone help me?! I need to know how to solve this problem before class!
 

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