Differential equations - exact equations

In summary, the conversation discusses finding the solution to an initial value problem involving a differential equation. The process involves determining whether the equation is exact and then finding a function f(t,y). There is a small mistake made in the process, but the final solution is y(t)^3t^2= 1 with a possible solution of y= t^-2/3. The conversation also mentions that this is a separable equation.
  • #1
braindead101
162
0
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.
 
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  • #2
You've made one silly little error- and it isn't "differential equations", it's basic algebra- maybe an arithmetic error!

braindead101 said:
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 [itex]2ty^3+ 3t^2y^2dy/dt= 0[/itex] which you can write in "differential form" as [itex](2ty^3)dt+ (3t^3y^2)dy= 0[/itex].

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, [tex]df= \frac{\partial f}{\partial t}dt+ \frac{\partial f}{\partial y}dy[/tex]

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

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,
[tex]\frac{\partial^2 f}{\partial y\partial t}= \frac{\partial (2ty^3}{\partial y}= 6ty^2[/tex]
and
[tex]\frac{\partial^2 f}{\partial t\partial y}= \frac{\partial (3t^2y^2}{\partial t}= 6ty^2[/tex]

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 [itex]\frac{\partial f}{\partial y}[/itex] 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 [itex]3t^2y^2[/itex] 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 [itex]f(t,y)= y^3t^2+ c[/itex]. 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: [itex]y^3t^2= C[/itex]


sub y(1) = 1

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

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 [itex]y= ^3\sqrt{1/t^2}= t^{-\frac{2}{3}}[/itex] 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 [itex]y^3t^2= 1[/itex] with respect to t, we get [itex]3y^2t^2 y'+ 2y^3t= 0[/itex]. Yes, that's exactly the original equation.

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

By the way, in addition to being exact, this is also a "separable" equation: we can write [itex]2ty^3 + 3t^2y^2 dy/dt = 0[/itex] as [itex]3t^2y^2 dy/dt= -2ty^3[/itex] and then, dividing both sides by [itex]y^3t^2[/itex], get [itex]3dy/y= -2dt/t[/itex]. Integrating both sides, 3 ln(y)= -2ln(t)+ C so
[itex]ln(y^3)= ln(t^{-2})+ C[/itex], [itex]y^3= Ct^{-2}[/itex] and, finally, [itex]t^2y^3= C[/itex] as before.
 
  • #3
oh wow, thanks so much
didnt correct that, so use to crossing things out and writing 1 afterwards..
 
  • #4
i mean didnt see that mistake.. lol
i did correct it
thanks a lot.
 

1. What is a differential equation?

A differential equation is a mathematical equation that relates the rates of change of a function to the function itself. It describes how a function changes over time or in relation to other variables.

2. What is an exact equation?

An exact equation is a type of differential equation where the terms can be manipulated to form an equivalent equation, making it easier to solve. This is achieved by finding an integrating factor that cancels out the non-exact terms.

3. How do you know if a differential equation is exact?

A differential equation is exact if it can be written in the form M(x,y) + N(x,y)y' = 0, where M and N are functions of x and y. The equation is exact if and only if ∂M/∂y = ∂N/∂x.

4. What are some common applications of exact equations in science?

Exact equations are used in many areas of science, such as physics, chemistry, and biology. They are particularly useful in modeling natural phenomena that involve rates of change, such as population growth, chemical reactions, and electrical circuits.

5. What are some techniques for solving exact equations?

There are several techniques for solving exact equations, including the method of separation of variables, substitution, and using an integrating factor. It is also important to understand how to apply initial conditions or boundary conditions to find specific solutions to the equation.

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