Pro. #31- GR0568

Nameless · Post by **Nameless** » Sat Oct 11, 2008 1:36 am

Hi dears,

Can we solve the ODE equation : y'=1+y^4 ?

thanks

lime · Post by **lime** » Sat Oct 11, 2008 4:34 am

My answer is
y = i[(1+c*exp(2ix)) / (1-c*exp(2ix))], where "c" is constant.

amateur · Post by **amateur** » Sat Oct 11, 2008 8:46 am

Thanks lime,

I tried to verify your answer. This is what I got:

dy/dx = - 4 c exp(2ix) / [ -1 + c exp(2ix) ]^2

while

y^4 + 1 = 2 [1 + 6 c^2 exp(4ix) + c^4 exp(8ix)] / [-1 + c exp(2ix) ]^4

I tried it the following way (So far am not done with the solution)

y^4 + 1 = (y + i^2)(y - i^2)
= [y + (1 + i)/sqrt(2)] [y - (1 + i)/sqrt(2)] [y + (1 - i)/sqrt(2)] [y - (1 - i)/sqrt(2)]

So resolve 1/(y^4 + 1) into partial fractions and then integrate each linear factor separately.

Finally, to answer this question (Q. 31) on the GRE, we don't need to solve this differential equation. Observe that the dy/dx = m = slope of the solution at various values of y.

For example at y = 0, dy/dx = m = 1 => 45 degrees
This rules out B, E.

At y = 1, -1, dy/dx = m = 2. So the slope of the curve should be symmetric on either side of the origin. This rules out D.

Finally at y = +infinity, -infinity, the slope should be tan 90 => 90 degrees. This rules out C.

The only answer left is A.

amateur · Post by **amateur** » Sat Oct 11, 2008 9:08 am

SORRY, THERE WERE SOME MINOR ERRORS IN THE LAST POST. THIS IS THE CORRECTED VERSION.

Thanks lime,

I tried to verify your answer. This is what I got:

dy/dx = - 4 c exp(2ix) / [ 1 - c exp(2ix) ]^2

while

y^4 + 1 = 2 [1 + 6 c^2 exp(4ix) + c^4 exp(8ix)] / [1 - c exp(2ix) ]^4

I tried it the following way (So far am not done with the solution)

The equation after separation of variables becomes

dy/ [y^4 + 1] = dx

y^4 + 1 = (y^2 + i)(y^2 - i)
= [y + (1 + i)/sqrt(2)] [y - (1 + i)/sqrt(2)] [y + (1 - i)/sqrt(2)] [y - (1 - i)/sqrt(2)]

So resolve 1/(y^4 + 1) into partial fractions and then integrate each linear factor separately. The final answer would be of the form x = ...

Finally, to answer this question (Q. 31) on the GRE, we don't need to solve this differential equation. Observe that the dy/dx = m = slope of the solution at various values of y.

For example at y = 0, dy/dx = m = 1 => 45 degrees
This rules out B, E.

At y = 1, -1, dy/dx = m = 2. So the slope of the curve should be symmetric on either side of the origin. This rules out D.

Finally at y = +infinity, -infinity, the slope should be tan 90 => 90 degrees. This rules out C.

The only answer left is A.

Nameless · Post by **Nameless** » Sat Oct 11, 2008 9:23 am

Thanks Lime, and amateur

@Amateur : I think the method which you proposed using partial factions works in this case

lime · Post by **lime** » Sat Oct 11, 2008 12:26 pm

My bad. I was solving another equation: y' = y^2+1

Sorry for confusing you.