differential equations - exact equations [Archive]

braindead101

Feb24-07, 05:23 AM

Find the solution to the initial value problem.
2ty^3 + 3t^2y^2 dy/dt = 0 , y(1) = 1

I found out whether the equation was exact and it was, and i continued as follows. NOTE: the answer is y(t) = t^(-4/3) i just don't know how to get it. or maybe my solution is just in another form.

im going to use f instead of that greek symbol i do not how to type out here.

M(t,y) = 2ty^3
N(t,y) = 3t^2y^2

don't know why, just copied from textbook.
M(t,y) = df(t,y)/dt = 2ty^3
N(t,y) = df(t,y)/dy = 3t^2y^2

f(t,y) = integ(M(t,y)dt) + h(y)
f(t,y) = integ(2ty^3 dt) + h(y)
f(t,y) = 2y^3(1/2t^2) + h(y)
f(t,y) = y^3t^2 + h(y)

df(t,y)/dy = t^2(3y^2) + dh(y)/dy
df(t,y)/dy = 3t^2y^2 + dh(y)/dy
3t^2y^2 = 3t^2y^2 + dh(y)/dy
dh(y)/dy = 1
h(y) = integ(1 dy) + c
h(y) = y + c

f(t,y) = y^3t^2 + y + c
y^3t^2 + y = C

sub y(1) = 1

(1)^3(1)^2 + 1 = C
C= 2
.'. y(t)^3t^2 + y(t) = 2

and i can't isolate y(t) so i left that as final solution. any thoughts on whether this is right or wrong.

HallsofIvy

Feb24-07, 06:32 AM

You've made one silly little error- and it isn't "differential equations", it's basic algebra- maybe an arithmetic error!

Find the solution to the initial value problem.
2ty^3 + 3t^2y^2 dy/dt = 0 , y(1) = 1

I found out whether the equation was exact and it was, and i continued as follows. NOTE: the answer is y(t) = t^(-4/3) i just don't know how to get it. or maybe my solution is just in another form.

im going to use f instead of that greek symbol i do not how to type out here.

M(t,y) = 2ty^3
N(t,y) = 3t^2y^2

don't know why, just copied from textbook.
Surely your textbook defines M and N?
Your differential equation is 2ty^3+ 3t^2y^2dy/dt= 0 which you can write in "differential form" as (2ty^3)dt+ (3t^3y^2)dy= 0.

You want to determine whether the left side is an "exact differential"- that is, if it can be written as df for some function f(t,y). By the chain rule, df= \frac{\partial f}{\partial t}dt+ \frac{\partial f}{\partial y}dy

So you want to know if there exist f such that
\frac{\partial f}{\partial t}= 2ty^3
and
\frac{\partial f}{\partial t}= 3t^2y^2

One way of determining whether such an f exists without finding it is to remember that the "mixed" derivatives are equal (as long as they are continuous). Here, if such an f exists,
\frac{\partial^2 f}{\partial y\partial t}= \frac{\partial (2ty^3}{\partial y}= 6ty^2
and
\frac{\partial^2 f}{\partial t\partial y}= \frac{\partial (3t^2y^2}{\partial t}= 6ty^2

Yes, those are the same so this equation is exact and such a function f(t,y) exists!

M(t,y) = df(t,y)/dt = 2ty^3
N(t,y) = df(t,y)/dy = 3t^2y^2

f(t,y) = integ(M(t,y)dt) + h(y)
f(t,y) = integ(2ty^3 dt) + h(y)
f(t,y) = 2y^3(1/2t^2) + h(y)
f(t,y) = y^3t^2 + h(y)

Good. In "reversing" partial integration with respect to t, in which you treat y as a constant, the "constant of integration might be a function of y- that's your "h(y)".

df(t,y)/dy = t^2(3y^2) + dh(y)/dy
df(t,y)/dy = 3t^2y^2 + dh(y)/dy
Strictly speaking, that should be \frac{\partial f}{\partial y} but that's fine. The dh/dy is an "ordinary" derivative since h depends only on y.
3t^2y^2 = 3t^2y^2 + dh(y)/dy
Yes, that partial derivative must be equal to the "N(t,y)" from the equation.

dh(y)/dy = 1
What? Is that a typo? Surely, subtracting 3t^2y^2 from both sides, dh/dy= 0!

h(y) = integ(1 dy) + c
h(y) = y + c
Since dh/dy= 0, h(y)= c, a constant. (And since h depends only on y, it really is a constant.)

f(t,y) = y^3t^2 + y + c
First, it should be f(t,y)= y^3t^2+ c. Second you don't really need the "c" here but it doesn't hurt. Since the equation is basically df= 0, f(t,y)= C and you can combine c and C- exactly as you do next.

y^3t^2 + y = C
Correction: y^3t^2= C

sub y(1) = 1

(1)^3(1)^2 + 1 = C
C= 2
.'. y(t)^3t^2 + y(t) = 2
With the correction (1^3)(1^2)= C so C= 1.
y^3t^2= 1

and i can't isolate y(t) so i left that as final solution. any thoughts on whether this is right or wrong.

Well, with the correction, you can write y= ^3\sqrt{1/t^2}= t^{-\frac{2}{3}} but you should be aware that, with first order differential equations you quite often can't solve for one variable as a function of the other. You could, for example, use "implicit differentiation" to see if the solution satisfies the differential equation: differentiating y^3t^2= 1 with respect to t, we get 3y^2t^2 y'+ 2y^3t= 0. Yes, that's exactly the original equation.

If we check you erroneous solution, y(t)^3t^2 + y(t) = 2 we get 3y^2t^2y'+ 2y^3t+ y'= 0 or 2y^3t+ (3y^2t^2+ 1)y'= 0 which is NOT the original equation.

By the way, in addition to being exact, this is also a "separable" equation: we can write 2ty^3 + 3t^2y^2 dy/dt = 0 as 3t^2y^2 dy/dt= -2ty^3 and then, dividing both sides by y^3t^2, get 3dy/y= -2dt/t. Integrating both sides, 3 ln(y)= -2ln(t)+ C so
ln(y^3)= ln(t^{-2})+ C, y^3= Ct^{-2} and, finally, t^2y^3= C as before.