1. ## 2 problems

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 dont even know how to start.

Thank you
B

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?
Given the family of curves

$
y=\frac{1}{x+C}
$
,

rearrange to give:

$
x+C=1/y
$

and differentiate:

$
1=-\frac{1}{y^2}\ \frac{dy}{dx}
$

or:

$
\frac{dy}{dx}=-y^2
$
.

So now we know that the slope of any curve of the family that passes
through $(x,y)$ is $-y^2$. So the slope of a curve orthogonal to a member
of the family passing through $(x,y)$ is $1/y^2$. So a curve in the orthoganal family
satisfies the ODE:

$
\frac{dy}{dx}=\frac{1}{y^2}
$

Which has solutions:

$
f(x,y)=\frac{y^3}{3}-x+K=0
$

Which is the required orthogonal family.

RonL

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 dont even know how to start.
Now you have me slightly confused here. As I understand things a pair
$\{f(x),g(x)\}$ is a base solution of $y''+p(x)y'+q(x)y=0$
means that $f$ and $g$ are linearly independent
(have non-zero Wronskian), and $Af(x)+Bg(x)$ is a general
solution of $y''+p(x)y'+q(x)y=0$.

Your pairs both can be used to give a general solution, and with a bit
of rearrangement can be seen to give the same solutions for suitable
choices of the multipliers.

RonL

4. Originally Posted by CaptainBlack
Now you have me slightly confused here. As I understand things a pair
$\{f(x),g(x)\}$ is a base solution of $y''+p(x)y'+q(x)y=0$
means that $f$ and $g$ are linearly independent
(have non-zero Wronskian), and $Af(x)+Bg(x)$ is a general
solution of $y''+p(x)y'+q(x)y=0$.

Your pairs both can be used to give a general solution, and with a bit
of rearrangement can be seen to give the same solutions for suitable
choices of the multipliers.

RonL
Yes you are right,
if you substract the two function that contitute the basis, I have the same general solution fot both sets:
Y=C1*x + C2 X^2.

thank you for both solutions.
B