Initial Value problem

Rowallan

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Joined
Mar 3, 2011
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11
I am having a mental block with this one!

x.dy/dx = (e^3x) - y(1-3x) Where x>0 and y(1) = e^3

I started by dividing through by x;

dy/dx = (1/x . e^3x) - (y/x) + 2y

I can't see how to separate the variables or see an itegrating factor!

Any help appreciated.
 
You can do it this way and then use an integrating factor:

You divided by x already. I think you made a booboo. Where did the 2y come from?.

So, we get:

\(\displaystyle \frac{dy}{dx}=\frac{e^{3x}}{x}-y(\frac{1}{x}-3)\)

Rearrange:

\(\displaystyle \frac{dy}{dx}+y(\frac{1}{x}-3)=\frac{e^{3x}}{x}\)

Now, the integrating factor is \(\displaystyle e^{\int(\frac{1}{x}-3)dx}\)

Can you continue?.
 
Rowallan said:
Could you expand how you got that. :?

It appears I missed that 'x' on the left side. Please don't use the period for multiplication. '*' works better. Learning a little LaTeX would work even better.
 
I am curious about the . for multiplication. Where is this 'taught'?. It is confusing because it would make a multiplication look like a decimal.

Unfortunately, I have seen it before and have always wondered.
 
galactus said:
Where is this 'taught'?

South America?

I don't like periods as operators, either. I've seen confusion with this issue on these boards before, too.

EG:

3.4.2.1 = 7.14


 
OK, here is where I've got now:

The integrating factor, p =exp(integ(1/x + 3) dx
=exp(ln(x) + 3x) since x>0
=x + exp(3x)

Multiplying both sides of the equation by p gives:

(x + exp(3x))dy/dx + (x + exp(3X))(1/X + 3)y = (exp(3x)/x)(x + exp(3X))

(x + exp(3x))dy/dx + (3x + exp(3X)/x + 3exp93x))y = (exp(3x) + (exp(6X))/x)

Is it now correct to say:

d/dy((1/x + 3)y) = exp(3x)(1 + (exp(3X))/x)

then integrating both sides:

(1/x + 3)y = (1/3)exp(3x) + ((1/6)exp(6x))/x + (exp(3x)(ln(x)) + c

Where do I go now?
 
Starting with the integrating factor:

\(\displaystyle e^{\int(\frac{1}{x}-3)dx}=e^{ln(x)-3x}=e^{ln(x)}\cdot e^{-3x}=xe^{-3x}\)

Multiply by IF:

\(\displaystyle \frac{d}{dx}[yxe^{-3x}]=\frac{e^{3x}}{x}\cdot xe^{-3x}\)

\(\displaystyle \frac{d}{dx}[yxe^{-3x}]=1\)

Integrate:

\(\displaystyle xye^{-3x}=x+C\)

Divide by \(\displaystyle xe^{-3x}\) and factor:

\(\displaystyle y=e^{-3x}\left(1+\frac{C}{x}\right)\)

Now, use the IC to find C and you're done.
 
Thanks, I follow what you have done but the integrating factor, p =exp(integ(1/x + 3) dx not exp(integ(1/x - 3) dx.
I'll see what I can do now though.
Thanks
 
The integrating factor, p =exp(integ(1/x + 3) dx
=exp(ln(x) + 3x) since x>0
=(x)( exp(3x))

Multiplying both sides of the equation by p gives:

d/dy[(y)(x)exp(3x)] = (exp(3X))/x)((x)exp(3x) = (exp(3x))(exp(3x)) = exp(6x)

then integrating both sides:

(y)(x)exp(3x)= Integ exp(6x) dx = (1/6)exp(6x) + c

So

y = ((1/6)exp(6x) + c)/((x)exp(3x)) = (exp(3x))/(6x) + c/(x)exp(3x)

With y(1) = exp(3)

y(1) = ((1/6)exp(6) + c)/exp(3) = exp(3)/(6) + c/exp(3) = exp(3)

So

c/exp(3) = exp(3) - exp(3)/(6) = (5/6)exp(3)

c = ((5/6)exp(3))(exp(3)) = (5/6)exp(6)

Thus

y = exp(3)/(6) + (5)exp(6)/(6)exp(3)

Am I near now?
 
Rowallan said:
The integrating factor, p =exp(integ(1/x + 3) dx
=exp(ln(x) + 3x) since x>0
=(x)( exp(3x))

Multiplying both sides of the equation by p gives:

d/dy[(y)(x)exp(3x)] = (exp(3X))/x)((x)exp(3x) = (exp(3x))(exp(3x)) = exp(6x)

then integrating both sides:

(y)(x)exp(3x)= Integ exp(6x) dx = (1/6)exp(6x) + c

So

y = ((1/6)exp(6x) + c)/((x)exp(3x)) = (exp(3x))/(6x) + c/(x)exp(3x)

With y(1) = exp(3)

y(1) = ((1/6)exp(6) + c)/exp(3) = exp(3)/(6) + c/exp(3) = exp(3)

So

c/exp(3) = exp(3) - exp(3)/(6) = (5/6)exp(3)

c = ((5/6)exp(3))(exp(3)) = (5/6)exp(6)

Thus

y = exp(3)/(6) + (5)exp(6)/(6)exp(3)

Am I near now?

How come your 'y' is a constant now ??!!
 
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