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Benny

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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?

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?

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