Struggling with yet another DE problem -_-

[Benny]

I don't know what it is with me and maths, I always get stuck on questions which most people wouldn't find to be all that difficult. I'm having trouble working out the following.

Consider Clairaut's equation y = xy' + f(y').

a) Show that y = cx + f(c) is a solution to the given DE.

y' = c + 0 = c since f(c) = constant so f'(c) = 0.

RHS = xy' + f(y') = xc + f(c) = cx + f(c) = y = LHS as required.

b) Show that Clairaut's equation also has a solution in parametric form:

x = -f('t)
y = f(t) - tf'(t)

and show that this is a different solution from that in 'a' (Hint: consider dy/dx for both solutions) and is therefore a singular solution.

Firstly, I've typed up the equations as given. The equation x = -f('t) doesn't make sense to me. I'm thinking that it might be x = -f'(t) or x = -f(t'). I'll go with the former.

I don't really understand what the hint is telling me. Seeing as the original equation clearly has y as a function of x I'm thinking that I need to use the chain rule. dy/dx = y' = (dy/dt)(dt/dx).

Simple calculations give dy/dt = -tf''(t). Assuming x = -f'(t) then dx/dt = -t'f''(t) then dy/dx = (-tf''(t))(-t'f''(t)) = t't''(f''(t))^2.

So RHS = xy' + f(y')
= xt't''(f''(t))^2 + f(t't''(f''(t))^2)
= -t't''f'(t)(f''(t))^2 + f(t't''(f''(t))^2)

That's all I can get. I can't thinking of a way to verify that RHS = LHS of y = xy' + f(y') for these parametric equations. Any help would be really good thanks.

Edit: Nevermind, I managed the verification of the parametric solution part with some creative algebra. However, it relied on the assumption that when they said x = -f('t) they actually meant x = -f'(t)

Now I need to know how I can show that the solution is different from that of part 'a' where the solution was y = cx + f(y'). Can somebody help me with that?

Also, the next part of the question gives me some DEs to solve. I'm thinking that it requires some usage of parts a and b.

Q. Find the solution/s of y = xy' + 1 - log(y').

Comparing with y = xy' - log(y') I see that f(y') = -log(y') but I don't know what to do with it.

The next one y = xy' -(y')^3, same deal. I see that f(y') = -(y')^3 but again, I cannot figure out what to do. Can someone please help me with these?

 

[saltydog]

Hey Benny, probably got this one already. Took me a while to figure it out but interesting. I'll post it for the record in case anyone else is interested:
(and I can't get those numbers to line up neither)

For the equation:

<br /> y=xy^{&#039;}+f(y^{&#039;})\tag{1}<br />


Differentiating with respect to x:

y^{&#039;}=xy^{&#039;&#039;}+y^{&#039;}+f^{&#039;}(y^{&#039;})y^{&#039;&#039;}

Or:

y^{&#039;&#039;}(x+f^{&#039;}(y^{&#039;})

Then either:

<br /> y^{&#039;&#039;}=0 \qquad\tag{2}<br />

Or:

<br /> x+f^{&#039;}(y^{&#039;})=0 \tag{3}<br />


The solution of (2) is:

<br /> y(x)=ax+b \tag{4}<br />

Substituting this into (1), the general solution is then:

y(x)=ax+f(a)\tag{5}


Note that the general solution is a family of straight lines, i.e., constant slopes.

Now equations (1) and (3) constitute a set of parametric equations of x and y in terms of the parameter y^{&#039;}[/tex], that is:<br /> <br /> x=-f^{&amp;#039;}(\alpha)\tag{6}<br /> <br /> y=-\alpha f^{&amp;#039;}(\alpha)+f(\alpha)\tag{7}<br /> <br /> That these are in fact a solution to (1) is shown by taking the derivatives of each with respect to \alpha:<br /> <br /> \frac{dx}{d\alpha}=-f^{&amp;#039;&amp;#039;}(\alpha})<br /> <br /> \frac{dy}{d\alpha}=f^{&amp;#039;}(\alpha)-(\alpha f^{&amp;#039;&amp;#039;}(\alpha)+f^{&amp;#039;}(\alpha))<br /> <br /> and therefore:<br /> <br /> &lt;br /&gt; \frac{dy}{dx}=\alpha\tag{8}<br /> <br /> Substituting the expressions for x, y, and y&#039; into (1) yields:<br /> <br /> -\alpha f^{&amp;#039;}(\alpha)+f(\alpha)+\alpha f^{&amp;#039;}(\alpha)-f(\alpha)=0<br /> <br /> Thus showing that the parametric forms satisfy (1).<br /> <br /> That the parametric solutions respresens a solution that is not a special case of the general solution, i.e., is a singular solution is given by (8) which gives the derivative of the parametric solution which is dependent on alpha, that is, is not constant like the derivative of the general solution.<br /> <br /> Edit:<br /> <br /> I assume you know how to solve for the singular solution. You know, for:<br /> <br /> x=-f^{&amp;#039;}(c)<br /> <br /> Calculate the inverse of f^{&amp;#039;}:<br /> <br /> c=-\left[f^{&amp;#039;}\right]^{-1}(x)<br /> <br /> Then substitute into:<br /> <br /> y=cx+f(c)<br /> <br /> Or just eliminate \alpha between (6) and (7).<br /> <br /> Rock and roll.

 

[saltydog]

Guys, I'm just not yet satisfied with this. Here's the two equations Benny suggested and I hope you've done them by now cus' I don't want to get in trouble with Hall for doing your homework but really we should work them up a little more to gain insight into the nature of the singular solutions since these come up sometimes with non-linear equations.

<br /> \begin{align}<br /> y&amp;=xy^{&#039;}+\left[1-ln(y^{&#039;})\right] \\<br /> y&amp;=xy^{&#039;}-\left(y^{&#039;}\right)^3 <br /> \end{align}<br />

So we have a method of solving these and are to obtain both a general solution y_g(x) and a singular solution y_s(x). The singular solution is:

(a) not a special case of the general solution, and

(b) is, at each of its points, tangent to some element of the one-parameter family that is the general solution. A curve that's tangent to some element of a one-parameter family of curves is called an envelope of that family.

Well, I'd like to plot both a few members of the general solution and verify that the singular solution does indeed envelop them. Is that too much to ask?

Solution of (1):

f(u)=1-ln(u)

f^{&#039;}(u)=-\frac{1}{u}

Thus:

y_g(x)=cx+\left(1-ln(c)\right)

Now (from above):

x=-f^{&#039;}(c)=\frac{1}{c}\Rightarrow c=\frac{1}{x}

And so the singular solution is:

y_s(x)=2+ln(x)

The fist plot exhibits this. The singular solution is a logarithmic curve and the two members of the general family (straight lines) are tangent to it. Others would be also.

Solution of (2):

f(u)=-u^3

f^{&#039;}(u)=-3u^2

Thus:

y_g(x)=cx-c^3

So that:

x=-f^{&#039;}(c)=3c^2\Rightarrow c=\pm \sqrt{\frac{x}{3}}

Thus we have two singular solutions:

y_{s1}(x)=\frac{2}{3}\frac{x^{3/2}}{\sqrt{3}}

y_{s2}(x)=-\frac{2}{3}\frac{x^{3/2}}{\sqrt{3}}

The second plot shows both singular solutions curving from the origin as well as two general solutions which are tangent to them.