1st order Differential Equations: 2 possible particular solutions

By Stuart Moore on Monday, June 25, 2001 - 12:30 am :

Going through some past STEP questions as practice for next week. Question 8(i), STEP 2 2000:

Let y be the solution of the differential equation

dy /dx + 4x e^-x 2(y +3)^1/2 = 0 (x > = 0)

that satisfies the condition y =6 when x =0.

Separating the variables gives

y = (e^{- x^{2} + C})^{2} - 3

where C is the constant of integration. Putting in the values gives

9=(1+C)²

(1+C)=±3

C = 2 or -4

hence there are two particular solutions. Is this correct, or have I missed something? The next part of the question says show that

y \to 1

x \to \infty

which is true for C = 2 but not C = -4, so 2 is presumably the one they were looking for, but I cannot see anything to stop -4 being valid as well.

By Tim Martin on Monday, June 25, 2001 - 10:46 am :

As far as I can tell the problem is that in solving to get

y = (e^{- x^{2}} + c)^{2} - 3

you have to square the equation, and then when you later take the square root you introduce another (false) solution. If you follow the algebra through but don't square you get:

$(y + 3)^{1 / 2} = e^{- x^{2}} + c$ and then you can substitute in to get the value of $c$ . One point to notice here is that when evaluating the square root you only introduce the $\pm$ if you introduced the square root. In this case the square root was in the original expression, so it is the positive root which is required.

By Stuart Moore on Monday, June 25, 2001 - 11:06 am :

Cheers, that seems to explain it. I must remember to work out my constants of integration at the first opportunity ;-)