Differential equations origin graph

[helix999]

will the graph of the solution to the differential equation dy/dx=1+y^4 pass through origin?

My solution graph is not passing through origin. If it should pass, then how?

 

[Mark44]

What's your solution and how did you get it? This is a separable DE, but the after it is separated, the resulting integral is not an obvious one.

 

[helix999]

Yes it is a separable one.

dy/(1+y^4) = dx

my solution was 1/2(sqrt(2)) tan inverse [(y^2-1)/(root2)y] - 1/4(root2) log [(y^2+1-(root2)y)/(y^2+1+(root2)y)] = x + c

I guess this graph will not pass through origin because we don't know value of c, so at x=0 how can we find the value of y?

 

[helix999]

mutiply & divide by 2
1/2 (2/(1+y^4))

add & subtract y^2 and separate in 2 integrals
1/2 (y^2+1/(y^4+1)) dx - 1/2 (y^2-1/(y^4+1)) dx

taking y^2 common from numerator & denominator and cancelling it, we will get the integrals in the form of dx/x^2+a^2 and dx/x^2-a^2 , by using their respective formulas i got the above solution.

 

[HallsofIvy]

I think you need to reread the question! There is no such thing as "the" solution to a differential equation so it makes no sense to ask if the graph of "the" solution passes through the origin. You are correct that the general solution contains an unknown constant "c". It does NOT then follow that "the graph will not pass through origin because we don't know value of c". Not knowing the value of c means we do not know if it passes through the origin or not.

 

[HallsofIvy]

mutiply & divide by 2
1/2 (2/(1+y^4))

add & subtract y^2 and separate in 2 integrals
1/2 (y^2+1/(y^4+1)) dx - 1/2 (y^2-1/(y^4+1)) dx

taking y^2 common from numerator & denominator and cancelling it, we will get the integrals in the form of dx/x^2+a^2 and dx/x^2-a^2 , by using their respective formulas i got the above solution.

This makes no sense at all. I accept that you really mean "dy" and not "dx" but still- once you have $(y^2+ 1)/(y^4+1)- (y^2-1)/(y^4+ 1)$ what do you mean by "taking y² common from numerator and denominator and cancelling it? There is no common factor of y² to cancel. Surely you don't think that
$$\frac{y^2+ 1}{y^4+ 1}= \frac{1}{y^2+ 1}$$
by "cancelling" y²!

 

[helix999]

The question I have is

Which graph most closely represents the graph of a solution to the differential equation dy/dx=1+y^4?

I have 5 graphs as options. I have doubt in two. Both are same except one is passing thru origin & the other is not. The correct answer is the graph which is passing through origin.

This makes no sense at all. I accept that you really mean "dy" and not "dx" but still- once you have $(y^2+ 1)/(y^4+1)- (y^2-1)/(y^4+ 1)$ what do you mean by "taking y² common from numerator and denominator and cancelling it? There is no common factor of y² to cancel. Surely you don't think that
$$\frac{y^2+ 1}{y^4+ 1}= \frac{1}{y^2+ 1}$$
by "cancelling" y²!

I meant:

1/2 [y^2(1+1/y^2)/y^2(y^2+1/y^2)] dy - 1/2 [y^2(1-1/y^2)/y^2(y^2+1/y^2)] dy

and then we can write y^2+1/y^2 remaining in the denominator as (y-1/y)^2 + (sqrt(2))^2 for the first integral and (y+1/y)^2-(sqrt(2))^2 for the second integral.

And then we can substitute y-1/y as t in the first integral to solve integral and y+1/y as u to solve the second integral so that they will fit into the given formulas.

 

[gabbagabbahey]

You're probably not even required to solve the DE...I'm guessing that for each graph, you are supposed to compare the approximate slope of the graph (dy/dx)≈(Δy/Δx) to 1+(y0)^4 at certain points (x0,y0) along the graph...if you find points where (dy/dx) is drastically different from 1+(y0)^4 then you know that it cannot be a Solution of the DE.

 

[helix999]

You're probably not even required to solve the DE...I'm guessing that for each graph, you are supposed to compare the approximate slope of the graph (dy/dx)≈(Δy/Δx) to 1+(y0)^4 at certain points (x0,y0) along the graph...if you find points where (dy/dx) is drastically different from 1+(y0)^4 then you know that it cannot be a Solution of the DE.

Fine...i got it.