Bridging the Gap: Solving Two Problems

[brad sue]

Hi ,
I have two problem with which I am stuck:

1- Given the family of curves y=1/(x+C).
Find the family of orthogonal trajectories.

For this problem, I took the first derivative of y---> y'=-1/(x+C)^2. from here, I cannot find the value of C since (x+C)^2=-1/y' which is impossible.
What can I do here?

2- The pairs {x^2+x,x^2} and {x,x+2x^2} ( found by the Wronskian) are base solutions of the equation y''+p(x)y'+q(x)y=0.
Give the general solution of the equation.

Here I don't even know how to start.

Thank you
B

 

[bomba923]

1- Given the family of curves y=1/(x+C).
Find the family of orthogonal trajectories.

For this problem, I took the first derivative of y---> y'=-1/(x+C)^2. from here, I cannot find the value of C since (x+C)^2=-1/y' which is impossible.
What can I do here?

Using 'k' instead of 'C', (in other words, let C=k)
Let g(x) describe the family of orthagonal trajectories of y=f(x)=1/(x+k)

$$\begin{gathered}<br /> y = f\left( x \right) = \frac{1}{{x + k}} \Rightarrow \frac{{df}}{{dx}} = - \frac{1}{{\left( {x + k} \right)^2 }} \Rightarrow \hfill \\<br /> \frac{{dg}}{{dx}} = \left( {x + k} \right)^2 \Rightarrow g\left( x \right) = \int {\left( {x + k} \right)^2 dx} = \frac{{\left( {x + k}\right)^3 }}{3} \hfill \\ \end{gathered}$$

 

[HallsofIvy]

Hi ,
I have two problem with which I am stuck:

1- Given the family of curves y=1/(x+C).
Find the family of orthogonal trajectories.

For this problem, I took the first derivative of y---> y'=-1/(x+C)^2. from here, I cannot find the value of C since (x+C)^2=-1/y' which is impossible.
What can I do here?

Why do say that's impossible? It's certainly possible for y' to be negative isn't it?
It might be simpler, however, to reverse it: x+ C= 1/y so, differentiating both sides, 1= (-1/y²)y' and y'= -y². For the orthogonal trajectories you must have y'= 1/y².

2- The pairs {x^2+x,x^2} and {x,x+2x^2} ( found by the Wronskian) are base solutions of the equation y''+p(x)y'+q(x)y=0.
Give the general solution of the equation.

Here I don't even know how to start.

Thank you
B

I'm not certain what is meant by "base solutions" or in what sense those are "base solutions" but I suspect that the general solution is
y(x)= Cx²+ Dx.

 

[brad sue]

Why do say that's impossible? It's certainly possible for y' to be negative isn't it?
It might be simpler, however, to reverse it: x+ C= 1/y so, differentiating both sides, 1= (-1/y²)y' and y'= -y². For the orthogonal trajectories you must have y'= 1/y².


I'm not certain what is meant by "base solutions" or in what sense those are "base solutions" but I suspect that the general solution is
y(x)= Cx²+ Dx.

Thank you,
I was confused by the fact that a square ca not be negative. But as you pointed out , we are talking about derivative..
B