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.(adsbygoogle = window.adsbygoogle || []).push({});

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?

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Struggling with yet another DE problem -_-

**Physics Forums | Science Articles, Homework Help, Discussion**