calc iv question... differentials

[simms]

want to help me with a Calc IV question?

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

 

[CTrain]

I know x^2 = 2x. I'm the winNAR !!

 

[chuckywang]

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

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

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.

 

[xEDIT409]

I'm only in Calc 2, but I gave it a shot.

x= -4?

 

[EarthwormJim]

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

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

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?

 

[Syringer]

Should've asked me a year and a half ago, you're too late.

/bump

 

[jagec]

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.

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?

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

y' != y.

 

[chuckywang]

bump cause I solved it.

I win.

 

[Random Variable]

Originally posted by: chuckywang
bump cause I solved it.

I win.

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.

 

[chuckywang]

Originally posted by: Vespasian

Originally posted by: chuckywang
bump cause I solved it.

I win.

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.

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.

 

[Gibson486]

Originally posted by: Vespasian

Originally posted by: chuckywang
bump cause I solved it.

I win.

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.

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.

 

[Random Variable]

Originally posted by: chuckywang

Originally posted by: Vespasian

Originally posted by: chuckywang
bump cause I solved it.

I win.

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.

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)

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.

 

[maziwanka]

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

 

[Random Variable]

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:

 

[chuckywang]

.

 

[Hankerton]

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.

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.

Woah chuckywang, do you teach calculus or something?

 

[Random Variable]

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

 

[m4ch0dude]

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

 

[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

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.

 

[chuckywang]

Originally posted by: Hankerton
Woah chuckywang, do you teach calculus or something?

Nah, I'm still a senior in college. But I like math.

 

[Random Variable]

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

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.

It's cool.

So what's the next problem?