How to solve this differential equation Watch

claret_n_blue
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#1
Report Thread starter 8 years ago
#1
dy/dt = 3y + 1 y(0)=0

When I integrate, I get:

1/(9y²) = t + c

But then you can't put the 0's in to find a value for c. What do you do?
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marcusmerehay
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#2
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You're integrating your LHS incorrectly.

\int \frac{1}{y} dy = ln(y) + constant.

Your solution should become

 t = \frac{1}{3} ln(3y+1)

Or equivalently

y=\frac{1}{3} (e^{3t}-1)
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claret_n_blue
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(Original post by marcusmerehay)
You're integrating your LHS incorrectly.

\int \frac{1}{y} dy = ln(y) + constant
Oh yeah!!! Duhhh. Lol I was differentiating.

Thanks :
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Gemini92
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(Original post by claret_n_blue)
dy/dt = 3y + 1 y(0)=0

When I integrate, I get:

1/(9y²) = t + c

But then you can't put the 0's in to find a value for c. What do you do?
You haven't rearranged/integrated correctly. You should be integrating 1/(3y+1) w.r.t y, which will have ln in it.
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