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Integration question

Find the solution y = y(x) of the following initial value problem:

dydx=x2y2\frac {dy}{dx} = x^2y^2 , y(−1) = 2

I separate it and get ∫1y2dy=∫x2dx\int \frac {1}{y^2}dy = \int x^2 dx

I know how to integrate I just don't know what my limits are. x is from 0 to x, but what is y? where does the y(-1) come into this?

Also what does it mean "Find the solution y = y(x)"
(edited 13 years ago)
Reply 1
Avatar for Pheylan
Pheylan
G A B R I E L
I know how to integrate I just don't know what my limits are. x is from 0 to x, but what is y? where does the y(-1) come into this?

integrate it without limits, then substitute x=-1 and y=2 to find the constant
G A B R I E L
Find the solution y = y(x) of the following initial value problem:

dydx=x2y2\frac {dy}{dx} = x^2y^2 , y(−1) = 2

I separate it and get ∫1y2dy=∫x2dx\int \frac {1}{y^2}dy = \int x^2 dx

I know how to integrate I just don't know what my limits are. x is from 0 to x, but what is y? where does the y(-1) come into this?

Also what does it mean "Find the solution y = y(x)"

Don't worry about the limits yet. Just integrate as normal and don't forget the arbitrary constant.
y(-1)=2 is telling you that y=2 when x=-1 and you need to use this after integration to solve for the constant of integration.
By 'Find the solution of y=y(x)', it's asking you to find the particular function (note that this is different to the general function) y of x which satisfies this differential equation given that the curve y(x) passes through the point (-1,2).
EDIT: Beaten to it. :p:
(edited 13 years ago)
Reply 3
Avatar for G A B R I E L
G A B R I E L
OP
Oh ok thanks I get y=3x2−0.5y = \frac {3}{x^2 - 0.5}

thanks
G A B R I E L
Oh ok thanks I get y=3x2−0.5y = \frac {3}{x^2 - 0.5}

thanks

That is not correct. Try again. If you can't spot what you've done wrong, post your working.
Reply 5
Avatar for G A B R I E L
G A B R I E L
OP
Farhan.Hanif93
That is not correct. Try again. If you can't spot what you've done wrong, post your working.


\frac {dy}{dx} = x^2y^2 , y(−1) = 2

I separate it and get

∫1y2dy=∫x2dx\int \frac {1}{y^2}dy = \int x^2 dx

=−1y=13x3+C = -\frac {1}{y} = \frac {1}{3}x^3 + C

If I let y = 2 and x = -1 then

=−12=−13+C = -\frac {1}{2} = -\frac {1}{3} + C

So C = −16-\frac {1}{6}

From here its simplification so either my simplification was wrong or my above was wrong.
G A B R I E L
\frac {dy}{dx} = x^2y^2 , y(−1) = 2

I separate it and get

∫1y2dy=∫x2dx\int \frac {1}{y^2}dy = \int x^2 dx

=−1y=13x3+C = -\frac {1}{y} = \frac {1}{3}x^3 + C

If I let y = 2 and x = -1 then

=−12=−13+C = -\frac {1}{2} = -\frac {1}{3} + C

So C = −16-\frac {1}{6}

From here its simplification so either my simplification was wrong or my above was wrong.

Something's gone wrong with your simplification. You had no x3x^3 term in your solution, which immediately suggests that it was wrong. Working through it mentally, I also think there's other problems with your solution. Post your simplification.

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