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calc iv question... differentials

Originally posted by: simms
want to help me with a Calc IV question?

xy'+2y=3x, y(1)=5 ... solve the dif eqn.

🙂
Click to expand...

Remember that the general solution to a diffeq is the homogenous solution + (any) particular solution.

The particular solution is y1(x) = x just be judicious guessing.

The homogenous solution is the solution to xy' + 2y = 0, which is the same as y' + 2/x*y = 0, which can be solved by variation of parameters (I assume you are on this topic now). Sparing you the work, the homogeneous solution is y2(x) = A*x^-2 where A is any constant.

Therefore, the entire solution is y(x) = y1(x) + y2(x) = A*x^-2 + x.

Since you know that y(1) = 5, you can solve for A by plugging in:

5 = A + 1 which means A = 4.

Your final solution is y(x) = 4x^-2 + x.
 
Originally posted by: simms
want to help me with a Calc IV question?

xy'+2y=3x, y(1)=5 ... solve the dif eqn.

🙂
Click to expand...

Probably wrong, but if Y(1)=5 doesn't that mean that at y=1, x=5? So shouldn't it just be plug and chug? Or is that too easy?

5y'+2(1)=3(5) y'=13/5? at (1,5)

meh?
 
Originally posted by: EarthwormJim
Originally posted by: simms
want to help me with a Calc IV question?

xy'+2y=3x, y(1)=5 ... solve the dif eqn.

🙂
Click to expand...

Probably wrong, but if Y(1)=5 doesn't that mean that at y=1, x=5? So shouldn't it just be plug and chug? Or is that too easy?

5y'+2(1)=3(5) y'=13/5? at (1,5)

meh?
Click to expand...

I believe the equation he is trying to convey is x*dy/dx + 2y=3x, y(1)=5.

y' != y.
 
Originally posted by: chuckywang
bump cause I solved it. 🙂

I win.
Click to expand...

I did it differently:


y'+ (2/x)y = 3 (divide each term by x)


P(x)= 2/x
Q(x)= 3

So the integrating factor is e^(Integral P(x)dx)= e^[2ln(x)] = x^2

(x^2)y' + 2xy = 3x^2
(x^2y)' = 3x^2
x^2y = x^3
y = x + C

5 = 1 + C
C = 4

So the final answer should be y(x)= x+4

In this particular problem you could have almost guessed the answer without going through all of the steps.
 
Originally posted by: Vespasian
Originally posted by: chuckywang
bump cause I solved it. 🙂

I win.
Click to expand...


So the final answer should be y(x)= x+4

In this particular problem you could have almost guessed the answer without going through all of the steps.
Click to expand...

Your answer doesn't work. Just plug it in:

xy' + 2y = x*1 + 2*(x+4) = 3x+8, which does not equal 3x.


Trust me, my answer is the only solution to the diffeq.
 
Originally posted by: Vespasian
Originally posted by: chuckywang
bump cause I solved it. 🙂

I win.
Click to expand...

I did it differently:


y'+ (2/x)y = 3 (divide each term by x)


P(x)= 2/x
Q(x)= 3

So the integrating factor is e^(Integral P(x)dx)= e^[2ln(x)] = x^2

(x^2)y' + 2xy = 3x^2
(x^2y)' = 3x^2
x^2y = x^3
y = x + C

5 = 1 + C
C = 4

So the final answer should be y(x)= x+4

In this particular problem you could have almost guessed the answer without going through all of the steps.
Click to expand...



yup, that is the "differential equation way"....

Since this is a calc class.....you need to do it the "calc way". I tried the differential equation way in my class, and i was marked wrong b/c we could not use it.

The y(1)=5 is the initial condition. You need this because you will first have to take the integral of the equation to get rid of dy and dx. AFter you take the integral, you will have to find the constant, which is why the y(1)=5 is there.

chuckywang solution is right, but I find it kind of trivial in yoru case b/c the P(x) method is not introduced until you get to a real differential equations class. This method, from inspection, needs the p(x) method b/c it is not Odinary differential equation, but a Partial Differential equation. ODE's are seperable, this PDE is not.

 
Originally posted by: chuckywang
Originally posted by: Vespasian
Originally posted by: chuckywang
bump cause I solved it. 🙂

I win.
Click to expand...


So the final answer should be y(x)= x+4

In this particular problem you could have almost guessed the answer without going through all of the steps.
Click to expand...

1) Your answer doesn't work. Just plug it in:

xy' + 2y = x*1 + 2*(x+4) = 3x+8, which does not equal 3x.



2)
Originally posted by: Vespasian

So the integrating factor is e^(Integral P(x)dx)
Click to expand...

The integrating factor is e^(-Integral P(x)dx). You forgot the -1 factor before the integral. Common mistake. 🙂


Trust me, my answer is the only solution to the diffeq.
Click to expand...

😱 😱
 
i didn't remember any diff eq, but solved it this way:

xy'+2y=0 is the homogenous solution.

divide by y and x to get:

(y'/y)=-2/x

(y'/y)=(ln(y))', so:

(ln(y))'=-2/x

which is:

ln(y)=-2ln(x)+C
y=Cx^-2

so the total solution is: y=Cx^-2+x (the particular solution was easy to get and you can see chuckywangs solution).

using the IC:

5=C+1 -> C=1

therefore:

y=4x^-2+x

 
Let me try again!

y'+ (2/x)y = 3 (divide each term by x)


P(x)= 2/x
Q(x)= 3

So the integrating factor is e^-(Integral P(x)dx)= e^-[2ln(x)] = -x^2

(-x^2)y' - 2(x)y = -3x^2
[(-x^2)y]' = -3x^2
(-x^2)y = -x^3 +C
y = x - C(x^-2)

5 = 1 - C(1)
C = -4

therefore y(x)= x + 4(x^-2)

w00t!!! :thumbsup:
 
Originally posted by: chuckywang
Originally posted by: simms
want to help me with a Calc IV question?

xy'+2y=3x, y(1)=5 ... solve the dif eqn.

🙂
Click to expand...

Remember that the general solution to a diffeq is the homogenous solution + (any) particular solution.

The particular solution is y1(x) = x just be judicious guessing.

The homogenous solution is the solution to xy' + 2y = 0, which is the same as y' + 2/x*y = 0, which can be solved by variation of parameters (I assume you are on this topic now). Sparing you the work, the homogeneous solution is y2(x) = A*x^-2 where A is any constant.

Therefore, the entire solution is y(x) = y1(x) + y2(x) = A*x^-2 + x.

Since you know that y(1) = 5, you can solve for A by plugging in:

5 = A + 1 which means A = 4.

Your final solution is y(x) = 4x^-2 + x.
Click to expand...


Woah chuckywang, do you teach calculus or something?
 
My first method was correct but I forgot to divide C by x^2. The integrating factor is negative only if you use the formula directly. Stop confusing me! 🙁

y'+ (2/x)y = 3 (divide each term by x)


P(x)= 2/x
Q(x)= 3

So the integrating factor is e^(Integral P(x)dx)= e^[2ln(x)] = x^2

(x^2)y' + 2xy = 3x^2
[(x^2)y]' = 3x^2
(x^2)y = x^3 + C
y = x + C/x^2

5 = 1 + C
C = 4

therefore y(x)= x + 4/x^2
 
Cliff notes...

This is the reason I bought a ti-89 when I did this stuff... and now I forgot how to make the calculator do it
 
Originally posted by: Vespasian
My first method was correct but I forgot to divide C by x^2. The integrating factor is negative only if you use the formula directly. Stop confusing me! 🙁

y'+ (2/x)y = 3 (divide each term by x)


P(x)= 2/x
Q(x)= 3

So the integrating factor is e^(Integral P(x)dx)= e^[2ln(x)] = x^2

(x^2)y' + 2xy = 3x^2
[(x^2)y]' = 3x^2
(x^2)y = x^3 + C
y = x + C/x^2

5 = 1 + C
C = 4

therefore y(x)= x + 4/x^2
Click to expand...

I realize that you were indeed correct about the integrating factor. I apologize for my mistake and I edited my previous responses to your posts. It seems that I got confused of what exactly the "integrating factor" is defined as.
 
Originally posted by: chuckywang
Originally posted by: Vespasian
My first method was correct but I forgot to divide C by x^2. The integrating factor is negative only if you use the formula directly. Stop confusing me! 🙁

y'+ (2/x)y = 3 (divide each term by x)


P(x)= 2/x
Q(x)= 3

So the integrating factor is e^(Integral P(x)dx)= e^[2ln(x)] = x^2

(x^2)y' + 2xy = 3x^2
[(x^2)y]' = 3x^2
(x^2)y = x^3 + C
y = x + C/x^2

5 = 1 + C
C = 4

therefore y(x)= x + 4/x^2
Click to expand...

I realize that you were indeed correct about the integrating factor. I apologize for my mistake and I edited my previous responses to your posts. It seems that I got confused of what exactly the "integrating factor" is defined as.
Click to expand...

It's cool.

So what's the next problem? 😛
 
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