Calculus  Seperable Equations
posted by COFFEE .
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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