Tuesday
July 22, 2014

Homework Help: Calculus - Seperable Equations

Posted by COFFEE on Tuesday, June 26, 2007 at 12:17pm.

Solve the separable differential equation (dy/dx)=y(1+x) for y and find the exact value for y(.3).

dy/dx = y(1+x)
dy/y = (1+x)dx
Integral (dy/y) = Integral (1+x)dx
ln (y) = x + (1/2)x^2 + C
y = e^(x + (1/2)x^2 + C)
y(0.3) = e^(0.345 + C)

I am stuck here. How do I solve for C to get the exact value?

Thanks!


y = e^(x + (1/2)x^2 + C)
(0.3) = e^.3 * e^.045 * e^C

e^c= .3/(e^.345)

c= ln ( .3/(e^.345) )


y = e^(x + (1/2)x^2 + C)
(0.3) = e^.3 * e^.045 * e^C

e^c= .3/(e^.345)

c= ln ( .3/(e^.345) )

So I plug in 0.3 for both x and y?

The exact value of y(0.3) would be:

y(0.3)=e^((0.3)+(1/2)(0.3)^2+(ln(0.3/0.345)))

Is this correct? Thanks.

OOps, I copied your work without thinking.

y = e^(x + (1/2)x^2 + C)

now, to solve for c, you have to have some boundry value, ie, the y for a given x. Then you solve for c.

Sorry, I wasn't thinking.

That's ok...I forgot to post the condition of dy/dx=y(1+x), y(0)=1...sorry!

So it would be:

y = e^(x + (1/2)x^2 + C)
1 = e^(0 + (1/2)0^2 + C)
1 = e^C
ln(1) = C
C = 0

Then...y = e^(x + (1/2)x^2)
y(0.3) = e^(0.3 + (1/2)(0.3^2))
y(0.3) = e^0.345

And that's it, correct?

Yes.

Thank You!

help me solve this problem..

solve for W: 2L+2W=38

please and thank you!

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